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Use the limit definition to show that $g'(0)$ exists but $g'(0)\neq \lim_{x\to 0} g'(x)$, where

$$g(x)=\begin{cases}x^2\sin\frac1x,&\text{when }x\neq0\\\\ 0,&\text{when }x=0\end{cases}$$

I find that when $x\neq 0$, $g'(x)=2x \sin\dfrac1x-\cos\dfrac1x$.

My problem is that when I can't compute

$$\lim_{x\to0}g'(x)=2\lim_{x\to0}x \sin\frac1x-\lim_{x\to0}\cos\frac1x$$

question from Rogawski, Jon. Calculus Single Variable. 2nd ed. New York: W.H. Freeman, 2012. Print.

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    I think a mistake I made is I didn't use the limit definition yet, rather I directly computed g'(x) at $x\neq0$.2012-11-04
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    I’m pretty sure that you’re expected to use ordinary differentiation formulas to handle $g'(x)$ for non-zero $x$; it’s only at $0$ that you need the limit definition. But for that you will have to go back to the difference quotient at $0$: $$g'(0)=\lim_{h\to 0}\frac{g(0+h)-g(0)}h\;.$$2012-11-04
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    Information on $lim_{x\to0}g'(x)$ http://www.wolframalpha.com/input/?i=limit+2xsin(1%2fx)-cos(1%2fx)2012-11-04
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    Using Brian's equation, I find $lim_{x\to0}g'(x)=0$2012-11-04
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    Yes: $h\to 0$, and $\sin\frac1h$ is bounded, so the product $\to0$.2012-11-04
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    If $lim_{x\to0}g'(x)=0$ and $g'(0)=0$, how could $g'(0)\neq \lim_{x\to 0} g'(x)$?2012-11-04
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    Sorry: I misread what you wrote in your previous comment. Everything that I’ve written has been aimed at the problem of showing that $g'(0)$ exists (and in fact is $0$); I’ve said nothing about $\lim\limits_{x\to 0}g'(x)$. *That* doesn’t exist.2012-11-04

2 Answers 2

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Here’s a pretty large for showing that $g'(0)$ exists. By definition

$$g'(0)=\lim_{h\to 0}\frac{g(0+h)-g(0)}h=\lim_{h\to0}\frac{g(h)}h=\lim_{h\to 0}\frac{h^2\sin\frac1h}h=\lim_{h\to 0}h\sin\frac1h\;;\tag{1}$$

can you evaluate that last limit?

For the rest, you already have

$$\lim_{x\to0}g'(x)=2\lim_{x\to0}x \sin\frac1x-\lim_{x\to0}\cos\frac1x\;.$$

The first limit on the righthand side is the same as the limit in $(1)$, and since $\cos\frac1x$ oscillates between $1$ and $-1$ infinitely often as $x\to 0$, the second limit on the righthand side doesn’t exist. But as N.S. already pointed out, if $\lim_{x\to0}g'(x)$ existed, so would

$$\lim_{x\to0}\cos\frac1x=2\lim_{x\to0}x \sin\frac1x-\lim_{x\to0}g'(x)\;;$$

since it doesn’t, neither does $\lim_{x\to0}g'(x)t$.

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    yeah it's 0.http://www.wolframalpha.com/input/?i=limit+h+sin+1%2Fh2012-11-04
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    It seems to me that $g'(0)=\lim_{x\to 0} g'(x)$ so I can't prove $g'(0)\neq \lim_{x\to 0} g'(x)$2012-11-04
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    @Raindrop: How can they be equal when one of them is $0$ and the other **doesn’t exist**?2012-11-04
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    The limit $\lim_{x\to0}g'(x)=2\lim_{x\to0}x \sin\frac1x-\lim_{x\to0}\cos\frac1x$ doesn't exist but the limit $g'(0)=\lim_{h\to 0}\frac{g(0+h)-g(0)}h=\lim_{h\to0}\frac{g(h)}h=\lim_{h\to 0}\frac{h^2\sin\frac1h}h=\lim_{h\to 0}h\sin\frac1h\;;\tag{1}$ exists2012-11-04
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    There's a chance that the question is incorrect because the textbook has a typo2012-11-04
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    @Raindrop: No, there isn’t: the question is correct. And your previous comment is exactly the point: $g'(0)$ exists and is $0$, but $\lim_{x\to0}g'(x)$ does not exist. Therefore they aren’t equal.2012-11-04
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    Thanks! When finding the limit, we use the equation of $g(x)$ for $x\neq0$ but when using limit definition, we use $g(0)=0$ while computing $g'(0)$.2012-11-04
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    @Raindrop: Now you’ve got it. And in case you were wondering, the point of this problem is that even though $g'(x)$ exists everywhere, it’s not continuous at $x=0$. This is an example of a function that is differentiable, but not *continuously* differentiable.2012-11-04
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Assume by contradiction that $\lim_{x \to 0} g'(x)$ exists. Then

$$\lim_{x\to0}g'(x)-2\lim_{x\to0}x \sin(1/x)$$

also exists, thus $\lim_{x \to 0} \cos(\frac{1}{x})$ exists.

Alternatelly, find two different sequences, $x_n, y_n$ so that

$$ \lim_n g'(x_n)=0$$ $$\lim_n g'(y_n)=1$$