Suppose $$ f(x) = \left\{\begin{array}{cr} e^{-1/x^2} & x \neq 0\\ 0 & x = 0 \end{array} \right. $$ Show that $f^\prime(0)$ exists and is equal to $0$, also verify that $f^{\prime\prime}$ exists and is equal to $0$.
Do I solve this by finding the left and right side limits of $f(x)$ as $x$ approaches $0$?
We can solve this problem by directly using the definition of the derivative as follows $$f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}$$ implies $$f'(0)=\lim_{x\rightarrow 0}\frac{e^{-\frac{1}{x^2}}}{x}=0.$$ The same to $f''(0)$.
Note sure if you have seen this limit definition of the derivative, let me know if you want me to explain further.
Note that $$ f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} $$
So let's look at $f'(0)$:
$$ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{e^{-1/x^2}}{x} $$
Now you have to evaluate this limit properly and the such - since another guy answered before I clicked submit I'll add a little bit more information:
In order to evaluate the limit I would suggest using L'Hôpital's rule. By this rule we know
$$ \lim_{x \to x_0} \frac{f(x)}{g(x)} = \lim_{x \to x_0} \frac{f'(x)}{g'(x)} $$
given that $\lim_{x \to x_0} f(x) = \lim_{x \to x_0} g(x) = 0$ or $\pm \infty$ and $\lim_{x \to x_0} \frac{f'(x)}{g'(x)}$ exists where $g'(x) \neq 0 \; \forall \; x \neq 0$.
It appears your limit follows these conditions, so it appears you can use this theorem.
