Prove $f(x)=1/x$ not differentiable at $x = 0$. Prove $f(x)=\frac{1}{x}$ not differentiable at $x = 0$.
Suppose $f(x)$ differentiable at $x=0$, then by definition 
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ 
exists.  But this means that 
$$\lim_{h\to 0}\frac{\frac{1}{x+h}-\frac{1}{0}}{h}$$ 
exists, however $\frac{1}{0}$ is not defined so then the limit does not exist, which leads to a contradiction.
Is this correct or is there a better and more rigourous way to prove this? 
 A: Since $x=0$ is not in the domain of $f(x)$, you can immediately assume that $f(x)$ will not be differentiable at $x=0$. However, you can still show that $f(x)=\frac{1}{x}$ is not differentiable at $x=0$ using the limit definition:
$\lim\limits_{h\to0} \dfrac{f(x+h)-f(x)}{h} = \lim\limits_{h\to0} \dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h} = \lim\limits_{h\to0} \dfrac{\dfrac{x}{x(x+h)}-\dfrac{x+h}{x(x+h)}}{h} = \lim\limits_{h\to0} \dfrac{-h}{x(x+h)}\cdot\dfrac{1}{h} = $
$\lim\limits_{h\to0} \dfrac{-1}{x(x+h)} = \dfrac{-1}{x^2}$.
Since $f'(x) = \dfrac{-1}{x^2}$ is not defined at $x=0$, $f(x)$ is not differentiable at $x=0$.
A: For a derivative to be existing, it need to be continuous in the first place.To be continuous, left and right limit should be equal to the value at that point. Now it is not defined at 0, meaning not continuous hence not differentiable. Also we can show as below that left and right limit are not same and hence discontinuous. 
$f(x^+)=\lim\limits_{h \to 0}f(x+h)=\dfrac{1}{x+h}$ for $x>0$
$f(0^+)=\dfrac{1}{h}>0$
but
$f(x^-)=\lim\limits_{h \to 0}f(x-h)=\dfrac{1}{x-h}<0$
$f(0^-)=\dfrac{1}{-h}<0$
A: The easiest way would be observing that the domain of $f(x)=\frac{1}{x}$ excludes $x=0$. If we evaluate $f(x)$ at $x=0$ then we get $f(0)=\frac{1}{0}$, which is undefined. Alternatively, you could recognize that the function $f(x)=\frac{1}{x}$ is only differentiable at $x=0$ if
$$ f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$
exists. Evaluating the limit at $x=0$ forms
\begin{align}f'(0)&=\lim_{h\to 0}\frac{f(h)-f(0)}{h}\\&=\lim_{h\to 0}\frac{1/h-f(0)}{h}\end{align} 
which doesn't exist because
$$f(0)=\frac{1}{0}$$
doesn't exist and because
$$\lim_{h\to 0}\frac{1}{h^2}$$
also doesn't exist.
