Example showing $limlimits_{x to x_0} xf(x) neq x_0limlimits_{x to x_0} f(x)$
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I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$
For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.
Can someone provide an example to this statement?
calculus algebra-precalculus limits examples-counterexamples discontinuous-functions
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up vote
2
down vote
favorite
I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$
For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.
Can someone provide an example to this statement?
calculus algebra-precalculus limits examples-counterexamples discontinuous-functions
1
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
– AlkaKadri
Nov 28 at 0:45
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
– Xander Henderson
Nov 28 at 0:45
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$
For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.
Can someone provide an example to this statement?
calculus algebra-precalculus limits examples-counterexamples discontinuous-functions
I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$
For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.
Can someone provide an example to this statement?
calculus algebra-precalculus limits examples-counterexamples discontinuous-functions
calculus algebra-precalculus limits examples-counterexamples discontinuous-functions
edited Nov 28 at 0:49
Servaes
22.2k33793
22.2k33793
asked Nov 28 at 0:41
Squaring the Circle is Easy
585
585
1
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
– AlkaKadri
Nov 28 at 0:45
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
– Xander Henderson
Nov 28 at 0:45
add a comment |
1
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
– AlkaKadri
Nov 28 at 0:45
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
– Xander Henderson
Nov 28 at 0:45
1
1
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
– AlkaKadri
Nov 28 at 0:45
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
– AlkaKadri
Nov 28 at 0:45
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
– Xander Henderson
Nov 28 at 0:45
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
– Xander Henderson
Nov 28 at 0:45
add a comment |
4 Answers
4
active
oldest
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up vote
3
down vote
accepted
HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
$$lim_{xto x_0}xf(x)
=left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
=x_0lim_{xto x_0}f(x),$$
so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.
1
Existence of the limit is enough for the product rule; continuity is not required.
– Y. Forman
Nov 28 at 0:49
@Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
– Servaes
Nov 28 at 0:54
add a comment |
up vote
1
down vote
Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.
On the right hand side $lim_{x to x_0} f(x)$ is not defined.
add a comment |
up vote
1
down vote
How about $x_0 = 0$, $f(x) = 1/x^2$?
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1
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If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$
So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
$$lim_{xto x_0}xf(x)
=left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
=x_0lim_{xto x_0}f(x),$$
so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.
1
Existence of the limit is enough for the product rule; continuity is not required.
– Y. Forman
Nov 28 at 0:49
@Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
– Servaes
Nov 28 at 0:54
add a comment |
up vote
3
down vote
accepted
HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
$$lim_{xto x_0}xf(x)
=left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
=x_0lim_{xto x_0}f(x),$$
so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.
1
Existence of the limit is enough for the product rule; continuity is not required.
– Y. Forman
Nov 28 at 0:49
@Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
– Servaes
Nov 28 at 0:54
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
$$lim_{xto x_0}xf(x)
=left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
=x_0lim_{xto x_0}f(x),$$
so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.
HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
$$lim_{xto x_0}xf(x)
=left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
=x_0lim_{xto x_0}f(x),$$
so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.
edited Nov 28 at 0:51
answered Nov 28 at 0:45
Servaes
22.2k33793
22.2k33793
1
Existence of the limit is enough for the product rule; continuity is not required.
– Y. Forman
Nov 28 at 0:49
@Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
– Servaes
Nov 28 at 0:54
add a comment |
1
Existence of the limit is enough for the product rule; continuity is not required.
– Y. Forman
Nov 28 at 0:49
@Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
– Servaes
Nov 28 at 0:54
1
1
Existence of the limit is enough for the product rule; continuity is not required.
– Y. Forman
Nov 28 at 0:49
Existence of the limit is enough for the product rule; continuity is not required.
– Y. Forman
Nov 28 at 0:49
@Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
– Servaes
Nov 28 at 0:54
@Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
– Servaes
Nov 28 at 0:54
add a comment |
up vote
1
down vote
Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.
On the right hand side $lim_{x to x_0} f(x)$ is not defined.
add a comment |
up vote
1
down vote
Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.
On the right hand side $lim_{x to x_0} f(x)$ is not defined.
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.
On the right hand side $lim_{x to x_0} f(x)$ is not defined.
Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.
On the right hand side $lim_{x to x_0} f(x)$ is not defined.
answered Nov 28 at 0:45
Siong Thye Goh
97.6k1463116
97.6k1463116
add a comment |
add a comment |
up vote
1
down vote
How about $x_0 = 0$, $f(x) = 1/x^2$?
add a comment |
up vote
1
down vote
How about $x_0 = 0$, $f(x) = 1/x^2$?
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up vote
1
down vote
up vote
1
down vote
How about $x_0 = 0$, $f(x) = 1/x^2$?
How about $x_0 = 0$, $f(x) = 1/x^2$?
answered Nov 28 at 0:45
Ethan Bolker
40.6k545107
40.6k545107
add a comment |
add a comment |
up vote
1
down vote
If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$
So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$
add a comment |
up vote
1
down vote
If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$
So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$
add a comment |
up vote
1
down vote
up vote
1
down vote
If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$
So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$
If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$
So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$
answered Nov 28 at 0:47
Y. Forman
11.4k523
11.4k523
add a comment |
add a comment |
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1
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
– AlkaKadri
Nov 28 at 0:45
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
– Xander Henderson
Nov 28 at 0:45