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.
 A: 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 $y_n$.
A: 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.
A: 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.
