# Is the function $f(x)=\sin(1/x)$ differentiable at $x=0$?

The function $f$ is defined by $f(x)= \sin(1/x)$ for any $x\neq 0$. For $x=0$, $f(x)=0$. Determine if the function is differentiable at $x=0$.

I know that it isn't differentiable at that point because $f$ is not continuous at $x=0$, but I need to prove it and I'm not sure how to use

$$m(a)= \lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$

with a piecewise function.

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LaTeX tip: Most function names are supposed to be upright, and you can get the common ones by preceding it with a backslash, e.g. \sin(x) instead of sin(x). If it is not built-in, you can use the \mathrm command, e.g. \mathrm{lcm}(a,b). You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. – Zev Chonoles Mar 8 '13 at 17:22
So just prove that it is not continous at $0$. – 1015 Mar 8 '13 at 17:22
or consider the difference quotient for $a=0$ at $x=\frac1{(2n+\frac12)\pi}$ – Hagen von Eitzen Mar 8 '13 at 17:26

To show that the function is not differentiable at $0$, you need to show that the limit $$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0}\frac{\sin(\frac 1x)}{x}$$ does not exist.

This can be done by finding two sequences $x_n$ and $y_n$ that both go to zero, but such that $$\frac{\sin (\frac {1}{x_n})}{x_n} \text{ and } \frac{\sin (\frac {1}{y_n})}{y_n}$$ have different limits as $n \to \infty$.

As Hagen von Eitzen mention in his comment, trying reciprocals of multiples of $\pi$ should help you find appropriate sequences $x_n$ and $y_n$.

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You need to prove that differentiability implies continuity.
Hint: consider the following limit
$$\lim_{x\to a}(f(x)-f(a))$$ and try to prove that it's equal to $0$ assuming that function $f$ is differentiable in point $a$, which means $f'(a)$ exists.

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My approach: Consider the left and right limits.
i.e. right limit when x --> 0+ and left limit when x --> 0-.

Also, we may consider y = 1/x, and somehow "convert" the limit when x --> 0+ to become the limit when y --> infinity. Similarly, "convert" the limit when x --> 0- to the limit when y --> -infinity. These two limits should be different.

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