By using the definition of limit only, prove that $lim_{xrightarrow 0} frac1{3x+1} = 1$
up vote
1
down vote
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By using the definition of a limit only, prove that
$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$
We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$
Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct
calculus limits epsilon-delta
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
asked Aug 14 '17 at 21:50
Ben Jones
18611
add a comment |
up vote
1
down vote
favorite
By using the definition of a limit only, prove that
$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$
We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$
Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct
calculus limits epsilon-delta
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
asked Aug 14 '17 at 21:50
Ben Jones
18611
The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
By using the definition of a limit only, prove that
$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$
We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$
Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct
calculus limits epsilon-delta
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
asked Aug 14 '17 at 21:50
Ben Jones
18611
By using the definition of a limit only, prove that
$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$
We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$
Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct
calculus limits epsilon-delta
calculus limits epsilon-delta
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
asked Aug 14 '17 at 21:50
Ben Jones
18611
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
asked Aug 14 '17 at 21:50
Ben Jones
18611
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
edited Nov 27 at 12:15
Martin Sleziak
44.6k7115269
44.6k7115269
asked Aug 14 '17 at 21:50
Ben Jones
18611
asked Aug 14 '17 at 21:50
Ben Jones
18611
asked Aug 14 '17 at 21:50
Ben Jones
18611
18611
The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59
add a comment |
The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59
The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59
The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.
if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.
$$frac14 < 3x+1< frac74$$
$$left|frac{1}{3x+4} right| < 4$$
Hence $$12delta < epsilon$$
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
add a comment |
up vote
0
down vote
Note that
$$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$
Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that
$$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$
Finally, given any $epsilon>0$,
$$left|frac{1}{1+3x}-1right|<epsilon$$
whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
– Mark Viola
Aug 18 '17 at 16:00
add a comment |
up vote
0
down vote
No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.
If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
$x > -frac 29$.
So, $|x| < frac 29 implies 3x+1 > frac 13 implies
left|dfrac {1}{3x+1}right|<3$.
So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
< |9x|$
To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.
So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.
Hence we define $delta = min(frac 19epsilon, frac 29)$
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.
if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.
$$frac14 < 3x+1< frac74$$
$$left|frac{1}{3x+4} right| < 4$$
Hence $$12delta < epsilon$$
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
add a comment |
up vote
2
down vote
accepted
If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.
if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.
$$frac14 < 3x+1< frac74$$
$$left|frac{1}{3x+4} right| < 4$$
Hence $$12delta < epsilon$$
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.
if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.
$$frac14 < 3x+1< frac74$$
$$left|frac{1}{3x+4} right| < 4$$
Hence $$12delta < epsilon$$
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.
if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.
$$frac14 < 3x+1< frac74$$
$$left|frac{1}{3x+4} right| < 4$$
Hence $$12delta < epsilon$$
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
edited Aug 14 '17 at 22:06
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
answered Aug 14 '17 at 21:59
Siong Thye Goh
97.4k1463116
97.4k1463116
add a comment |
add a comment |
up vote
0
down vote
Note that
$$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$
Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that
$$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$
Finally, given any $epsilon>0$,
$$left|frac{1}{1+3x}-1right|<epsilon$$
whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
– Mark Viola
Aug 18 '17 at 16:00
add a comment |
up vote
0
down vote
Note that
$$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$
Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that
$$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$
Finally, given any $epsilon>0$,
$$left|frac{1}{1+3x}-1right|<epsilon$$
whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
– Mark Viola
Aug 18 '17 at 16:00
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that
$$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$
Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that
$$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$
Finally, given any $epsilon>0$,
$$left|frac{1}{1+3x}-1right|<epsilon$$
whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
Note that
$$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$
Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that
$$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$
Finally, given any $epsilon>0$,
$$left|frac{1}{1+3x}-1right|<epsilon$$
whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
answered Aug 14 '17 at 21:59
Mark Viola
129k1273170
129k1273170
Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
– Mark Viola
Aug 18 '17 at 16:00
add a comment |
Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
– Mark Viola
Aug 18 '17 at 16:00
Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
– Mark Viola
Aug 18 '17 at 16:00
Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
– Mark Viola
Aug 18 '17 at 16:00
add a comment |
up vote
0
down vote
No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.
If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
$x > -frac 29$.
So, $|x| < frac 29 implies 3x+1 > frac 13 implies
left|dfrac {1}{3x+1}right|<3$.
So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
< |9x|$
To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.
So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.
Hence we define $delta = min(frac 19epsilon, frac 29)$
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
add a comment |
up vote
0
down vote
No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.
If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
$x > -frac 29$.
So, $|x| < frac 29 implies 3x+1 > frac 13 implies
left|dfrac {1}{3x+1}right|<3$.
So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
< |9x|$
To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.
So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.
Hence we define $delta = min(frac 19epsilon, frac 29)$
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
add a comment |
up vote
0
down vote
up vote
0
down vote
No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.
If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
$x > -frac 29$.
So, $|x| < frac 29 implies 3x+1 > frac 13 implies
left|dfrac {1}{3x+1}right|<3$.
So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
< |9x|$
To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.
So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.
Hence we define $delta = min(frac 19epsilon, frac 29)$
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.
If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
$x > -frac 29$.
So, $|x| < frac 29 implies 3x+1 > frac 13 implies
left|dfrac {1}{3x+1}right|<3$.
So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
< |9x|$
To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.
So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.
Hence we define $delta = min(frac 19epsilon, frac 29)$
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
answered Aug 14 '17 at 23:21
steven gregory
17.6k32257
17.6k32257
add a comment |
add a comment |
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The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59