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Show that the function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $$f(x)=\begin{cases} e^{-1/x^2}\sin{\frac{1}{x}} & \text{if } ~x\ne 0 \\ 0 & \text{if}~~ x=0 \end{cases}$$ is differentiable and $f'$ is continuous at the point $x=0$.

attempt:

$f'(x)=\frac{e^{-1/x^2}}{x^3}(2 \sin{(1/x)}-x\cos{(1/x)})$

$f'(0)=\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0}\frac{1}{h}e^{-1/h^2}\sin{\frac{1}{h}}$

I am struck to proceed further. Please help with a detailed solution.

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    Show that $f'(0)$ exists, i.e. calculate that limit (you could use L'Hopital's rule, for example)...at any rate, it's obviously $0$ since the exponential is rapidly decaying compared to the $1/h$ and the $\sin(1/h)$ isn't doing anything. Then show that $f'(0) = \lim_{x \to 0} f'(x)$ by showing that the limit of your first expression as $x$ goes to $0$ is also $0$. (Same reasoning). – 2017-02-20
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    @hunter "at any ***rate***" HAHA I LOVE THE PUNS – 2017-02-20

1 Answers 1

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Note that

$$\left|\frac{e^{-1/h^2}\sin\frac1h}h\right|\le\frac{e^{-1/h^2}}{|h|}\to0$$

which follows from

$$\lim_{h\to0}\frac{e^{-1/h^2}}{|h|}=\lim_{x\to\infty}\frac{x}{e^{x^2}}$$

and apply L'Hospital's rule.

Same method to take $\lim_{x\to0}f'(x)$.

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    Please elaborate "Same method to take $\lim_{x\to0}f'(x)$" – 2017-02-20
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    By L Hospital Rule $$\lim_{x\to\infty}\frac{e^{-x^2}}{1/x}=\lim_{x\to\infty}2\frac{e^{-x^2}}{1/x^3}=\lim_{x\to\infty}\frac{4}{3}\frac{e^{-x^2}}{1/x^5}$$ again 0/0 form after repetition, how to get the limit zero? – 2017-02-20
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    Check the derivative of the numerator. – 2017-02-20
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    Or check my update. – 2017-02-20