0
$\begingroup$

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.

  • 0
    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

1

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 $\lim_{x\to0}g'(x)t$.

  • 1
    @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
2

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