How to check if $f(x)=x^2sin(x)^{-1}$ is differentiable at 0?
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1
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How would I check if $f(x)=x^2sin(x)^{-1}$ is differentiable at 0?
I tried using the definition of a derivative but got stuck at the limit part. Does one care what happens at the limit point, or only what happens near it?
begin{align}
f^prime(0) &= lim_{xto0}dfrac{f(x)-f(0)}{x-0}\
&= lim_{xto0}dfrac{x^2sin(x)^{-1}-0}{x-0}\
&= lim_{xto0}dfrac{x}{sin(x)}\
&=1? text{ or does it not exist?}
end{align}
calculus derivatives
asked Nov 17 at 22:08
kaisa
717
add a comment |
up vote
1
down vote
favorite
How would I check if $f(x)=x^2sin(x)^{-1}$ is differentiable at 0?
I tried using the definition of a derivative but got stuck at the limit part. Does one care what happens at the limit point, or only what happens near it?
begin{align}
f^prime(0) &= lim_{xto0}dfrac{f(x)-f(0)}{x-0}\
&= lim_{xto0}dfrac{x^2sin(x)^{-1}-0}{x-0}\
&= lim_{xto0}dfrac{x}{sin(x)}\
&=1? text{ or does it not exist?}
end{align}
calculus derivatives
asked Nov 17 at 22:08
kaisa
717
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 at 22:14
@TedShifrin I agree it is an uncorrect way to write that but from the derivation presented it seems to be intepreted as $frac{x^2}{sin x}$. Indeed I've used that notation to indicate that.
– gimusi
Nov 17 at 22:34
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How would I check if $f(x)=x^2sin(x)^{-1}$ is differentiable at 0?
I tried using the definition of a derivative but got stuck at the limit part. Does one care what happens at the limit point, or only what happens near it?
begin{align}
f^prime(0) &= lim_{xto0}dfrac{f(x)-f(0)}{x-0}\
&= lim_{xto0}dfrac{x^2sin(x)^{-1}-0}{x-0}\
&= lim_{xto0}dfrac{x}{sin(x)}\
&=1? text{ or does it not exist?}
end{align}
calculus derivatives
asked Nov 17 at 22:08
kaisa
717
How would I check if $f(x)=x^2sin(x)^{-1}$ is differentiable at 0?
I tried using the definition of a derivative but got stuck at the limit part. Does one care what happens at the limit point, or only what happens near it?
begin{align}
f^prime(0) &= lim_{xto0}dfrac{f(x)-f(0)}{x-0}\
&= lim_{xto0}dfrac{x^2sin(x)^{-1}-0}{x-0}\
&= lim_{xto0}dfrac{x}{sin(x)}\
&=1? text{ or does it not exist?}
end{align}
calculus derivatives
calculus derivatives
asked Nov 17 at 22:08
kaisa
717
asked Nov 17 at 22:08
kaisa
717
asked Nov 17 at 22:08
kaisa
717
asked Nov 17 at 22:08
kaisa
717
asked Nov 17 at 22:08
kaisa
717
717
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 at 22:14
@TedShifrin I agree it is an uncorrect way to write that but from the derivation presented it seems to be intepreted as $frac{x^2}{sin x}$. Indeed I've used that notation to indicate that.
– gimusi
Nov 17 at 22:34
add a comment |
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 at 22:14
@TedShifrin I agree it is an uncorrect way to write that but from the derivation presented it seems to be intepreted as $frac{x^2}{sin x}$. Indeed I've used that notation to indicate that.
– gimusi
Nov 17 at 22:34
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 at 22:14
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 at 22:14
@TedShifrin I agree it is an uncorrect way to write that but from the derivation presented it seems to be intepreted as $frac{x^2}{sin x}$. Indeed I've used that notation to indicate that.
– gimusi
Nov 17 at 22:34
@TedShifrin I agree it is an uncorrect way to write that but from the derivation presented it seems to be intepreted as $frac{x^2}{sin x}$. Indeed I've used that notation to indicate that.
– gimusi
Nov 17 at 22:34
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Note that
$$lim_{xto 0} frac{x^2}{sin x}=lim_{xto 0} xcdot frac{x}{sin x} =0 $$
and since $f(x)$ is not defined at $x=0$ we need to define $f(0)=0$ to make it countinuous at that point. Then, as you have shown, since by the definition the limit exists, $f(x)$ is differentiable at $x=0$.
answered Nov 17 at 22:10
gimusi
88.9k74495
add a comment |
up vote
1
down vote
Doing a simple Taylor expansion works well:
When $x$ approches $0$, $frac{x}{sin(x)} = frac{x}{x - frac{x^3}{6}+text{o}(x^3)} = frac{1}{1 - frac{x^2}{6}+text{o}(x^2)} rightarrow 1$.
So, yeah
$
begin{align}
f^prime(0) &= lim_{xto0}dfrac{x}{sin(x)} =1.
end{align}
$
answered Nov 17 at 22:46
Euler Pythagoras
4389
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Note that
$$lim_{xto 0} frac{x^2}{sin x}=lim_{xto 0} xcdot frac{x}{sin x} =0 $$
and since $f(x)$ is not defined at $x=0$ we need to define $f(0)=0$ to make it countinuous at that point. Then, as you have shown, since by the definition the limit exists, $f(x)$ is differentiable at $x=0$.
answered Nov 17 at 22:10
gimusi
88.9k74495
add a comment |
up vote
2
down vote
accepted
Note that
$$lim_{xto 0} frac{x^2}{sin x}=lim_{xto 0} xcdot frac{x}{sin x} =0 $$
and since $f(x)$ is not defined at $x=0$ we need to define $f(0)=0$ to make it countinuous at that point. Then, as you have shown, since by the definition the limit exists, $f(x)$ is differentiable at $x=0$.
answered Nov 17 at 22:10
gimusi
88.9k74495
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Note that
$$lim_{xto 0} frac{x^2}{sin x}=lim_{xto 0} xcdot frac{x}{sin x} =0 $$
and since $f(x)$ is not defined at $x=0$ we need to define $f(0)=0$ to make it countinuous at that point. Then, as you have shown, since by the definition the limit exists, $f(x)$ is differentiable at $x=0$.
answered Nov 17 at 22:10
gimusi
88.9k74495
Note that
$$lim_{xto 0} frac{x^2}{sin x}=lim_{xto 0} xcdot frac{x}{sin x} =0 $$
and since $f(x)$ is not defined at $x=0$ we need to define $f(0)=0$ to make it countinuous at that point. Then, as you have shown, since by the definition the limit exists, $f(x)$ is differentiable at $x=0$.
answered Nov 17 at 22:10
gimusi
88.9k74495
answered Nov 17 at 22:10
gimusi
88.9k74495
answered Nov 17 at 22:10
gimusi
88.9k74495
answered Nov 17 at 22:10
gimusi
88.9k74495
88.9k74495
add a comment |
add a comment |
up vote
1
down vote
Doing a simple Taylor expansion works well:
When $x$ approches $0$, $frac{x}{sin(x)} = frac{x}{x - frac{x^3}{6}+text{o}(x^3)} = frac{1}{1 - frac{x^2}{6}+text{o}(x^2)} rightarrow 1$.
So, yeah
$
begin{align}
f^prime(0) &= lim_{xto0}dfrac{x}{sin(x)} =1.
end{align}
$
answered Nov 17 at 22:46
Euler Pythagoras
4389
add a comment |
up vote
1
down vote
Doing a simple Taylor expansion works well:
When $x$ approches $0$, $frac{x}{sin(x)} = frac{x}{x - frac{x^3}{6}+text{o}(x^3)} = frac{1}{1 - frac{x^2}{6}+text{o}(x^2)} rightarrow 1$.
So, yeah
$
begin{align}
f^prime(0) &= lim_{xto0}dfrac{x}{sin(x)} =1.
end{align}
$
answered Nov 17 at 22:46
Euler Pythagoras
4389
add a comment |
up vote
1
down vote
up vote
1
down vote
Doing a simple Taylor expansion works well:
When $x$ approches $0$, $frac{x}{sin(x)} = frac{x}{x - frac{x^3}{6}+text{o}(x^3)} = frac{1}{1 - frac{x^2}{6}+text{o}(x^2)} rightarrow 1$.
So, yeah
$
begin{align}
f^prime(0) &= lim_{xto0}dfrac{x}{sin(x)} =1.
end{align}
$
answered Nov 17 at 22:46
Euler Pythagoras
4389
Doing a simple Taylor expansion works well:
When $x$ approches $0$, $frac{x}{sin(x)} = frac{x}{x - frac{x^3}{6}+text{o}(x^3)} = frac{1}{1 - frac{x^2}{6}+text{o}(x^2)} rightarrow 1$.
So, yeah
$
begin{align}
f^prime(0) &= lim_{xto0}dfrac{x}{sin(x)} =1.
end{align}
$
answered Nov 17 at 22:46
Euler Pythagoras
4389
edited Nov 20 at 15:58
answered Nov 17 at 22:46
Euler Pythagoras
4389
answered Nov 17 at 22:46
Euler Pythagoras
4389
answered Nov 17 at 22:46
Euler Pythagoras
4389
4389
add a comment |
add a comment |
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@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 at 22:14
@TedShifrin I agree it is an uncorrect way to write that but from the derivation presented it seems to be intepreted as $frac{x^2}{sin x}$. Indeed I've used that notation to indicate that.
– gimusi
Nov 17 at 22:34