If $f(t)\in \mathcal{C}[-1,1]$ then evaluate $\lim_{h\to\infty} \frac{1}{h}\int_{-h}^hf(t)dt$. 
If $f(t)\in \mathcal{C}[-1,1]$ then evaluate 
  $$\lim_{h\to 0} \frac{1}{h}\int_{-h}^hf(t)dt$$

I have just used fundamental theorem of integral calculus. However, I could not estimate this...that is, I have considered that $g(t)=\int_{-h}^t f(x)dx$. So the $g'(0)=f(0)-f(-h)$ and by the standard way we get $g'(0)=\lim_{h\to 0}\frac{g(h)-g(0)}{h}$ and substituting $g(t)$ and simplifying gives $$f(0)-f(-h)=2\lim_{h\to 0}\int_{-h}^hf(t)dt+\lim_{h\to 0}\frac 1 h \int_0^hf(t)dt$$ nothing more than this. Kindly help me to evaluate this....                        
 A: Assuming $\mathcal{C}$ are the continuous functions and $h\rightarrow 0$, we can use the mean value theorem, so there exists $x\in [-h,h]$ such that the integral is equal to $2hf(x)$. Again, using continuity of $f$, the expression converges to $2f(0)$.
A: An other way to prove it:
I note $F(x)=\int_0^x f(t)dt$ and I'll suppose that $h\to 0$.
$$\frac{1}{h}\int_{-h}^h f(t)dt=\frac{F(h)-F(-h)}{h}=\frac{F(h)-F(0)}{h}-\frac{F(-h)-F(0)}{h}$$
Then 
$$\lim_{h\to 0}\frac{1}{h}\int_{-h}^hf(t)dt=\lim_{h\to 0}\frac{F(h)-F(0)}{h}-\frac{F(-h)-F(0)}{h}\underset{u=-h}{=}\lim_{h\to 0}\frac{F(h)-F(0)}{h}+\lim_{u\to 0}\frac{F(u)-F(0)}{u}=2F'(0)=2f(0).$$
A: using the substitution $t=hu$
$$
\lim_{h\to 0} \frac{1}{h}\int_{-h}^hf(t)dt = \lim_{h\to 0} \int_{-1}^1f(hu)du = 2f(0)
$$
since $f$ is continuous and bounded on $[-1,1]$
A: Note that : $$\frac{1}{h} \int_{-h}^h f(x) \ \mathrm{d}x =\frac{1}{ħ}\int_0^h f(x) \ \mathrm{d}x + \frac{1}{(-h)} \int_0^{(-h)} f(x) \ \mathrm{d}x$$ By the fundamental theorem of calculus ($f$ is continuous), the limit is $f(0)+f(0)=2f(0)$.
