Definition of the derivative $f(x) = x^{3}\cos{\left(\frac{1}{x^{2}}\right)}, x \not= 0, f(0) = 0$
Show, by definition of the derivative, that $f$ is differentiable at $x = 0$ and find the derivative of $f$ there.
So we know the derivative is defined as:
$$f'{(x)} = \lim_{h \to 0}{\frac{f{(x + h)} - f{(x)}}{h}}$$
So we have:
$$f'{(x)} = \lim_{h \to 0}{\frac{{(x + h)}^{3}\cos{\left(\frac{1}{(x+h)^{2}}\right)} - x^{3}\cos{\left(\frac{1}{x^{2}}\right)}}{h}}$$
How would I evaluate this limit?

Your definition is incorrect. The right definition is
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
With this definition, we get
$$f'(0) = \lim_{h \to 0} \frac{h^3 \cos (1/h^2)}{h}$$
Can you take it from here? H: Use L'hôpital's rule. EDIT: Forget L'hôpital, which does not apply in this situation! Just note that $|\cos(1/h^2)| \leq 1$ for all $h \neq 0$.
