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...
 A: For the first limit, the limit doesn't exist. It goes to $\infty$ from the right and $-\infty$ from the left.
For the updated second limit, we can solve it using Taylor's theorem. In particular, since your function is twice continuously differentiable, we know that
$$ f(x) = 0 + 3x + f''(\xi)\frac{x^2}{2}$$
for some $\xi \in (0, x)$. Further, since $f''(x)$ is continuous, and $f''(0) = 5$, as $x \to 0$, we know that $\xi \to 0$ and so $f''(\xi) \to 5$. In particular, restricting $x$ sufficiently close to $0$ forces $|f''(\xi) - 5 | < \epsilon$. Then
$$ \int_0^h f(x) dx \approx \int_0^h 3x + \frac{5x^2}{2} dx = \frac{3h^2}{2} + \frac{5h^3}{6},$$
where I've dropped any $\epsilon$ considerations because they don't ultimately matter. Dividing by $h^2$ and taking $h \to 0$ shows that the limit is $\frac{3}{2}$.
A: 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 $-\infty$)
A: We can use l'hospitals rule for both limits.
For the first limit we have:
$$\lim_{h\to0}\frac{f\left(h\right)+f\left(-h\right)}{h^{2}} =\lim_{h\to0}\frac{f^{\prime}\left(h\right)-f^{\prime}\left(-h\right)}{2h}
 =\lim_{h\to0}\frac{1}{2}\left(\frac{f^{\prime}\left(h\right)-f^{\prime}\left(0\right)}{h}+\frac{-f^{\prime}\left(-h\right)+f^{\prime}\left(0\right)}{h}\right)
 =\frac{1}{2}\left(\lim_{h\to0}\frac{f^{\prime}\left(h\right)-f^{\prime}\left(0\right)}{h}+\lim_{h\to0}\frac{-f^{\prime}\left(-h\right)+f^{\prime}\left(0\right)}{h}\right)$$
By definition of the derivative $\lim_{h\to0}\frac{f^{\prime}\left(h\right)-f^{\prime}\left(0\right)}{h}=f^{\prime\prime}\left(0\right)$,
and by substitution: $\lim_{h\to0}\frac{-f^{\prime}\left(-h\right)+f^{\prime}\left(0\right)}{h}=\lim_{h\to0}\frac{-f^{\prime}\left(h\right)+f^{\prime}\left(0\right)}{-h}=\lim_{h\to0}\frac{f^{\prime}\left(h\right)-f^{\prime}\left(0\right)}{h}=f^{\prime\prime}\left(0\right)$.
So our result is:$$\lim_{h\to0}\frac{f\left(h\right)+f\left(-h\right)}{h^{2}} =\frac{1}{2}\left(f^{\prime\prime}\left(0\right)+f^{\prime\prime}\left(0\right)\right)
 =f^{\prime\prime}\left(0\right)=5$$
Now for the second limit we use the fundamental theorem when differentiating the numerator:$$\lim_{h\to0}\frac{\int_{0}^{h}f\left(x\right)dx}{h^{2}}=\lim_{h\to0}\frac{f\left(h\right)-f\left(0\right)}{2h}=\lim_{h\to0}\frac{1}{2h}\left(\frac{f\left(h\right)-f\left(0\right)}{h}\right)$$
As stated above $\lim_{h\to0}\frac{f^{\prime}\left(h\right)-f^{\prime}\left(0\right)}{h}=f^{\prime\prime}\left(0\right)=5$,
and of course $\lim_{h\to0}\frac{1}{2h}=\infty$.
Now we can conclude: $$\lim_{h\to0}\frac{\int_{0}^{h}f\left(x\right)dx}{h^{2}}=\infty$$
