Question regarding specific limits I was going through some past final exams for my "Analysis 1" class and I came across the following problem, which I've been so far unable to solve.
Let $f''$ be continuous in $(-1,1)$; with $f(0)=0$, $f'(0)=3$ and $f''(0)=5$.
$$
\lim_{h\rightarrow 0}\frac{f(h)+f(-h)}{h^{2}}
\\
\lim_{h\rightarrow 0}\frac{1}{h^{2}}\int_{0}^{h}f(x)dx
$$
Regarding the first limit, I came up with he following, although I'm not sure it's correct:
Given that:
$$\lim_{h\rightarrow 0}\frac{f(h))}{h^{2}}=\lim_{h\rightarrow 0}\frac{f(-h))}{h^{2}}$$
This gives:
$$\lim_{h\rightarrow 0}\frac{f(h)+f(-h)}{h^{2}}=2\lim_{h\rightarrow 0}\frac{f(h)}{h^{2}}=2\lim_{h\rightarrow 0}\frac{f(0+h)-f(0)}{h} \frac{1}{h}=
2 \left [\lim_{h\rightarrow 0}\frac{f(0+h)-f(0)}{h} \ \lim_{h\rightarrow 0}\frac{1}{h}  \right ]$$
Which tends to $+\infty$
As I said, I'm not completely convinced by my reasoning. Confirmation/another solution would be helpful.
I haven't been able to solve the 2nd limit. Any ideas?
EDIT: I made a typo on the 2nd limit. Its $\frac{1}{h^{2}}$ rather than $h^{2}$. Sorry...

For the second limit, note that $f$ is continuous, and hence so is $|f|$. So, there exists $m,M$ such that $m \leq |f| \leq M$. So, by using these bounds, we can show that the second limit is $0$. The first limit does not exist, as $lim_{h \rightarrow 0}\frac{1}{h}$ does not exist. (for $h \rightarrow 0+$the limit is $\infty$, while for $h \rightarrow 0-$ the limit is $-\infty$)
