How can I prove this integral is equal to f(0)? Given that $f$ continuous over $[-1,1]$, how can I show 
$\lim_{x \to 0}\frac{1}{x}\int_0^xf(t)dt = f(0)$?
I know the limit of $\frac{1}{X}$ doesn't exist at 0, and it's negative infinity from the left and positive infinity from the right, but I'm not sure how that helps anything.
 A: This is actually the definition of a derivative, so there is no need for L'Hospital. Specifically, consider 
$$F(x)=\int_0^x f(t) dt.$$
Then by definition $F(0)=0$. So your expression can be written as $\frac{F(x)-F(0)}{x-0}$. So the limit is again by definition $F'(0)$. This derivative can be readily calculated using the Fundamental Theorem.
A: $\lim_{x \to 0}\frac{1}{x}\int_0^xf(t)dt = \lim_{x \to 0}\frac{\int_0^xf(t)dt-\int_0^0f(t)dt}{x-0}= \frac{d}{dx}\int_0^xf(t)dt |_{x=0} = f(x) |_{x=0} = f(0)$
A: If $f(t)$ is the constant $f(0)$ this is trivial. So write
$$f(t) = f(0) + g(t),$$
with $g(t) = f(t) - f(0).$ We have that $g$ is continuous and $g(0)=0.$
We need to show 
$$
\lim \limits_{x \to 0 } \frac{1}{x} \int\limits_{0}^{x} g(t) dt =0.
$$
Given $\epsilon >0,$ $\exists \delta >0,$ such that 
$|t| < \delta \implies |g(t) | < \epsilon.$ 
We then have for $|x| < \delta$
$$\left| \frac{1}{x} \int\limits_{0}^{x} g(t) dt \right | \leq |x| \frac{1}{|x|} \sup \limits_{|t| < \delta} |g(t)| < \epsilon.$$
This say precisely that the limit is zero as claimed.
