$limlimits_{xto 0} frac{tan x - sin x}{x^3}$?
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$$lim_{xto 0} frac{tan x - sin x}{x^3}$$
Solution
begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}
But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.
calculus limits proof-verification limits-without-lhopital
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up vote
6
down vote
favorite
$$lim_{xto 0} frac{tan x - sin x}{x^3}$$
Solution
begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}
But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.
calculus limits proof-verification limits-without-lhopital
7
$infty -infty = ?$
– Math Lover
Nov 21 at 17:33
edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37
The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38
But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44
4
Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36
|
show 6 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
$$lim_{xto 0} frac{tan x - sin x}{x^3}$$
Solution
begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}
But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.
calculus limits proof-verification limits-without-lhopital
$$lim_{xto 0} frac{tan x - sin x}{x^3}$$
Solution
begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}
But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.
calculus limits proof-verification limits-without-lhopital
calculus limits proof-verification limits-without-lhopital
edited 17 hours ago
amWhy
191k27223437
191k27223437
asked Nov 21 at 17:31
Raghav
437
437
7
$infty -infty = ?$
– Math Lover
Nov 21 at 17:33
edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37
The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38
But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44
4
Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36
|
show 6 more comments
7
$infty -infty = ?$
– Math Lover
Nov 21 at 17:33
edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37
The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38
But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44
4
Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36
7
7
$infty -infty = ?$
– Math Lover
Nov 21 at 17:33
$infty -infty = ?$
– Math Lover
Nov 21 at 17:33
edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37
edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37
The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38
The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38
But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44
But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44
4
4
Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36
Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36
|
show 6 more comments
5 Answers
5
active
oldest
votes
up vote
10
down vote
accepted
Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...
Only split an initial limit into a product if the individual limits are defined.
Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
– Raghav
Nov 21 at 17:39
1
You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
– KM101
Nov 21 at 17:40
Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
– Raghav
Nov 21 at 17:46
Exactly! (No problem.)
– KM101
Nov 21 at 17:47
@Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
– Paul Sinclair
Nov 22 at 2:58
|
show 2 more comments
up vote
27
down vote
This is just another way of saying what the others told you.
$$lim_{xto 0} frac{tan x - sin x}{x^3}
ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$
The theorem is
IF $displaystyle lim_{xto 0}f(x) = L$
and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$
But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.
This limit can be evaluated without resorting to L'Hospital.
begin{align}
frac{tan x - sin x}{x^3}
&= frac{frac{sin x}{cos x} - sin x}{x^3} \
&= frac{sin x - sin x cos x}{x^3 cos x} \
&= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac{2sin^2(frac 12x)}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
end{align}
which approaches $dfrac 12$ as $x$ approaches $0$.
1
@Steven where did I do $infty - infty$
– Akash Roy
2 days ago
1
Your way is correct.
– KM101
2 days ago
1
@AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
– steven gregory
2 days ago
1
I didn't know we needed to go to Le Hospital to solve limits :)
– Abraham Zhang
2 days ago
1
@stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
– Abraham Zhang
2 days ago
|
show 5 more comments
up vote
12
down vote
I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.
New contributor
2
This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
– MPW
Nov 21 at 18:30
There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
– Acccumulation
Nov 21 at 19:57
Is my answer okay?
– Akash Roy
2 days ago
add a comment |
up vote
3
down vote
Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$
$$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$
Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
– Akash Roy
2 days ago
Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
– Jean-Claude Arbaut
2 days ago
Ok sir Jean Claude Arbaut . Thanks for your words.
– Akash Roy
2 days ago
That's a nice solution.
– Raghav
2 days ago
Thanks @Raghav bro.
– Akash Roy
2 days ago
add a comment |
up vote
1
down vote
Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.
To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...
Only split an initial limit into a product if the individual limits are defined.
Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
– Raghav
Nov 21 at 17:39
1
You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
– KM101
Nov 21 at 17:40
Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
– Raghav
Nov 21 at 17:46
Exactly! (No problem.)
– KM101
Nov 21 at 17:47
@Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
– Paul Sinclair
Nov 22 at 2:58
|
show 2 more comments
up vote
10
down vote
accepted
Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...
Only split an initial limit into a product if the individual limits are defined.
Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
– Raghav
Nov 21 at 17:39
1
You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
– KM101
Nov 21 at 17:40
Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
– Raghav
Nov 21 at 17:46
Exactly! (No problem.)
– KM101
Nov 21 at 17:47
@Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
– Paul Sinclair
Nov 22 at 2:58
|
show 2 more comments
up vote
10
down vote
accepted
up vote
10
down vote
accepted
Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...
Only split an initial limit into a product if the individual limits are defined.
Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...
Only split an initial limit into a product if the individual limits are defined.
answered Nov 21 at 17:37
KM101
2,254415
2,254415
Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
– Raghav
Nov 21 at 17:39
1
You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
– KM101
Nov 21 at 17:40
Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
– Raghav
Nov 21 at 17:46
Exactly! (No problem.)
– KM101
Nov 21 at 17:47
@Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
– Paul Sinclair
Nov 22 at 2:58
|
show 2 more comments
Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
– Raghav
Nov 21 at 17:39
1
You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
– KM101
Nov 21 at 17:40
Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
– Raghav
Nov 21 at 17:46
Exactly! (No problem.)
– KM101
Nov 21 at 17:47
@Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
– Paul Sinclair
Nov 22 at 2:58
Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
– Raghav
Nov 21 at 17:39
Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
– Raghav
Nov 21 at 17:39
1
1
You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
– KM101
Nov 21 at 17:40
You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
– KM101
Nov 21 at 17:40
Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
– Raghav
Nov 21 at 17:46
Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
– Raghav
Nov 21 at 17:46
Exactly! (No problem.)
– KM101
Nov 21 at 17:47
Exactly! (No problem.)
– KM101
Nov 21 at 17:47
@Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
– Paul Sinclair
Nov 22 at 2:58
@Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
– Paul Sinclair
Nov 22 at 2:58
|
show 2 more comments
up vote
27
down vote
This is just another way of saying what the others told you.
$$lim_{xto 0} frac{tan x - sin x}{x^3}
ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$
The theorem is
IF $displaystyle lim_{xto 0}f(x) = L$
and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$
But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.
This limit can be evaluated without resorting to L'Hospital.
begin{align}
frac{tan x - sin x}{x^3}
&= frac{frac{sin x}{cos x} - sin x}{x^3} \
&= frac{sin x - sin x cos x}{x^3 cos x} \
&= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac{2sin^2(frac 12x)}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
end{align}
which approaches $dfrac 12$ as $x$ approaches $0$.
1
@Steven where did I do $infty - infty$
– Akash Roy
2 days ago
1
Your way is correct.
– KM101
2 days ago
1
@AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
– steven gregory
2 days ago
1
I didn't know we needed to go to Le Hospital to solve limits :)
– Abraham Zhang
2 days ago
1
@stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
– Abraham Zhang
2 days ago
|
show 5 more comments
up vote
27
down vote
This is just another way of saying what the others told you.
$$lim_{xto 0} frac{tan x - sin x}{x^3}
ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$
The theorem is
IF $displaystyle lim_{xto 0}f(x) = L$
and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$
But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.
This limit can be evaluated without resorting to L'Hospital.
begin{align}
frac{tan x - sin x}{x^3}
&= frac{frac{sin x}{cos x} - sin x}{x^3} \
&= frac{sin x - sin x cos x}{x^3 cos x} \
&= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac{2sin^2(frac 12x)}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
end{align}
which approaches $dfrac 12$ as $x$ approaches $0$.
1
@Steven where did I do $infty - infty$
– Akash Roy
2 days ago
1
Your way is correct.
– KM101
2 days ago
1
@AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
– steven gregory
2 days ago
1
I didn't know we needed to go to Le Hospital to solve limits :)
– Abraham Zhang
2 days ago
1
@stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
– Abraham Zhang
2 days ago
|
show 5 more comments
up vote
27
down vote
up vote
27
down vote
This is just another way of saying what the others told you.
$$lim_{xto 0} frac{tan x - sin x}{x^3}
ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$
The theorem is
IF $displaystyle lim_{xto 0}f(x) = L$
and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$
But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.
This limit can be evaluated without resorting to L'Hospital.
begin{align}
frac{tan x - sin x}{x^3}
&= frac{frac{sin x}{cos x} - sin x}{x^3} \
&= frac{sin x - sin x cos x}{x^3 cos x} \
&= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac{2sin^2(frac 12x)}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
end{align}
which approaches $dfrac 12$ as $x$ approaches $0$.
This is just another way of saying what the others told you.
$$lim_{xto 0} frac{tan x - sin x}{x^3}
ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$
The theorem is
IF $displaystyle lim_{xto 0}f(x) = L$
and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$
But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.
This limit can be evaluated without resorting to L'Hospital.
begin{align}
frac{tan x - sin x}{x^3}
&= frac{frac{sin x}{cos x} - sin x}{x^3} \
&= frac{sin x - sin x cos x}{x^3 cos x} \
&= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac{2sin^2(frac 12x)}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
end{align}
which approaches $dfrac 12$ as $x$ approaches $0$.
edited Nov 21 at 18:19
answered Nov 21 at 17:57
steven gregory
17.5k22257
17.5k22257
1
@Steven where did I do $infty - infty$
– Akash Roy
2 days ago
1
Your way is correct.
– KM101
2 days ago
1
@AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
– steven gregory
2 days ago
1
I didn't know we needed to go to Le Hospital to solve limits :)
– Abraham Zhang
2 days ago
1
@stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
– Abraham Zhang
2 days ago
|
show 5 more comments
1
@Steven where did I do $infty - infty$
– Akash Roy
2 days ago
1
Your way is correct.
– KM101
2 days ago
1
@AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
– steven gregory
2 days ago
1
I didn't know we needed to go to Le Hospital to solve limits :)
– Abraham Zhang
2 days ago
1
@stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
– Abraham Zhang
2 days ago
1
1
@Steven where did I do $infty - infty$
– Akash Roy
2 days ago
@Steven where did I do $infty - infty$
– Akash Roy
2 days ago
1
1
Your way is correct.
– KM101
2 days ago
Your way is correct.
– KM101
2 days ago
1
1
@AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
– steven gregory
2 days ago
@AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
– steven gregory
2 days ago
1
1
I didn't know we needed to go to Le Hospital to solve limits :)
– Abraham Zhang
2 days ago
I didn't know we needed to go to Le Hospital to solve limits :)
– Abraham Zhang
2 days ago
1
1
@stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
– Abraham Zhang
2 days ago
@stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
– Abraham Zhang
2 days ago
|
show 5 more comments
up vote
12
down vote
I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.
New contributor
2
This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
– MPW
Nov 21 at 18:30
There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
– Acccumulation
Nov 21 at 19:57
Is my answer okay?
– Akash Roy
2 days ago
add a comment |
up vote
12
down vote
I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.
New contributor
2
This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
– MPW
Nov 21 at 18:30
There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
– Acccumulation
Nov 21 at 19:57
Is my answer okay?
– Akash Roy
2 days ago
add a comment |
up vote
12
down vote
up vote
12
down vote
I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.
New contributor
I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.
New contributor
edited Nov 21 at 18:26
MPW
29.4k11856
29.4k11856
New contributor
answered Nov 21 at 17:55
Reinstein
1212
1212
New contributor
New contributor
2
This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
– MPW
Nov 21 at 18:30
There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
– Acccumulation
Nov 21 at 19:57
Is my answer okay?
– Akash Roy
2 days ago
add a comment |
2
This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
– MPW
Nov 21 at 18:30
There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
– Acccumulation
Nov 21 at 19:57
Is my answer okay?
– Akash Roy
2 days ago
2
2
This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
– MPW
Nov 21 at 18:30
This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
– MPW
Nov 21 at 18:30
There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
– Acccumulation
Nov 21 at 19:57
There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
– Acccumulation
Nov 21 at 19:57
Is my answer okay?
– Akash Roy
2 days ago
Is my answer okay?
– Akash Roy
2 days ago
add a comment |
up vote
3
down vote
Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$
$$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$
Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
– Akash Roy
2 days ago
Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
– Jean-Claude Arbaut
2 days ago
Ok sir Jean Claude Arbaut . Thanks for your words.
– Akash Roy
2 days ago
That's a nice solution.
– Raghav
2 days ago
Thanks @Raghav bro.
– Akash Roy
2 days ago
add a comment |
up vote
3
down vote
Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$
$$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$
Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
– Akash Roy
2 days ago
Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
– Jean-Claude Arbaut
2 days ago
Ok sir Jean Claude Arbaut . Thanks for your words.
– Akash Roy
2 days ago
That's a nice solution.
– Raghav
2 days ago
Thanks @Raghav bro.
– Akash Roy
2 days ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$
$$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$
Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$
$$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$
edited 2 days ago
answered 2 days ago
Akash Roy
53615
53615
Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
– Akash Roy
2 days ago
Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
– Jean-Claude Arbaut
2 days ago
Ok sir Jean Claude Arbaut . Thanks for your words.
– Akash Roy
2 days ago
That's a nice solution.
– Raghav
2 days ago
Thanks @Raghav bro.
– Akash Roy
2 days ago
add a comment |
Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
– Akash Roy
2 days ago
Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
– Jean-Claude Arbaut
2 days ago
Ok sir Jean Claude Arbaut . Thanks for your words.
– Akash Roy
2 days ago
That's a nice solution.
– Raghav
2 days ago
Thanks @Raghav bro.
– Akash Roy
2 days ago
Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
– Akash Roy
2 days ago
Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
– Akash Roy
2 days ago
Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
– Jean-Claude Arbaut
2 days ago
Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
– Jean-Claude Arbaut
2 days ago
Ok sir Jean Claude Arbaut . Thanks for your words.
– Akash Roy
2 days ago
Ok sir Jean Claude Arbaut . Thanks for your words.
– Akash Roy
2 days ago
That's a nice solution.
– Raghav
2 days ago
That's a nice solution.
– Raghav
2 days ago
Thanks @Raghav bro.
– Akash Roy
2 days ago
Thanks @Raghav bro.
– Akash Roy
2 days ago
add a comment |
up vote
1
down vote
Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.
To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.
add a comment |
up vote
1
down vote
Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.
To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.
add a comment |
up vote
1
down vote
up vote
1
down vote
Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.
To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.
Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.
To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.
edited Nov 22 at 4:39
answered Nov 22 at 4:26
AmbretteOrrisey
3568
3568
add a comment |
add a comment |
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7
$infty -infty = ?$
– Math Lover
Nov 21 at 17:33
edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37
The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38
But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44
4
Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36