Mixing
How to check if $f(x)=x^2sin(x)^{-1}$ is differentiable at 0?
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 '18 at 22:08
kaisa
1019
add a comment |
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 '18 at 22:08
kaisa
1019
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 '18 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 '18 at 22:34
add a comment |
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 '18 at 22:08
kaisa
1019
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 '18 at 22:08
kaisa
1019
asked Nov 17 '18 at 22:08
kaisa
1019
asked Nov 17 '18 at 22:08
kaisa
1019
asked Nov 17 '18 at 22:08
kaisa
1019
asked Nov 17 '18 at 22:08
kaisa
1019
1019
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 '18 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 '18 at 22:34
add a comment |
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 '18 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 '18 at 22:34
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 '18 at 22:14
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 '18 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 '18 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 '18 at 22:34
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 22:10
gimusi
1
add a comment |
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 '18 at 22:46
Euler Pythagoras
51310
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002879%2fhow-to-check-if-fx-x2-sinx-1-is-differentiable-at-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 22:10
gimusi
1
add a comment |
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 '18 at 22:10
gimusi
1
add a comment |
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 '18 at 22:10
gimusi
1
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 '18 at 22:10
gimusi
1
answered Nov 17 '18 at 22:10
gimusi
1
answered Nov 17 '18 at 22:10
gimusi
1
answered Nov 17 '18 at 22:10
gimusi
1
1
add a comment |
add a comment |
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 '18 at 22:46
Euler Pythagoras
51310
add a comment |
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 '18 at 22:46
Euler Pythagoras
51310
add a comment |
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 '18 at 22:46
Euler Pythagoras
51310
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 '18 at 22:46
Euler Pythagoras
51310
edited Nov 20 '18 at 15:58
answered Nov 17 '18 at 22:46
Euler Pythagoras
51310
answered Nov 17 '18 at 22:46
Euler Pythagoras
51310
answered Nov 17 '18 at 22:46
Euler Pythagoras
51310
51310
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002879%2fhow-to-check-if-fx-x2-sinx-1-is-differentiable-at-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
@TedShifrin I thought he was referring to $frac{x^2}{sin x}$.
– gimusi
Nov 17 '18 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 '18 at 22:34