Let $g$ be a function that is continuous at 0 and satisfies $g(0) \ne 0$. Set $f(x) = x^2/g(x)$.
1) Explain why the quotient rule does not apply to $f$. 2) Show that $f$ is differentiable at 0 and calculate $f'(0)$
For 1) I don't understand why the quotient rule does not apply to $f$.
I have:
$f'(x) = \frac{2xg(x) - x^2g'(x)}{g(x)^2}$
And I'm unsure how this shows that the quotient rule does not apply to $f$, I think I am good for 2). Any help would be appreciated thanks!
Because $g$ is not necessarily differentiable at $x=0$.
To compute that derivative of $f$ at $x=0$, just go through by definition: \begin{align*} \lim_{h\rightarrow 0}\dfrac{f(h)-f(0)}{h}&=\lim_{h\rightarrow 0}\dfrac{h^{2}/g(h)-0}{h}\\ &=\lim_{h\rightarrow 0}\dfrac{h}{g(h)}\\ &=\left(\lim_{h\rightarrow 0}h\right)\left(\lim_{h\rightarrow 0}\dfrac{1}{g(h)}\right)\\ &=0\cdot\dfrac{1}{g(0)}\\ &=0, \end{align*} where the continuity of $g$ at $x=0$ is used.