# 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.

• Try using L'Hospital's rule. Commented Jan 30, 2015 at 1:54
• We haven't done L'Hopital's rule yet, can it be done without it?
– CIM
Commented Jan 30, 2015 at 1:55
• Use L'Hospital's Rule since the quotient is an indeterminate form. Commented Jan 30, 2015 at 1:55
• This is actually the definition of a derivative, so there is no need for L'Hospital: you have $F(x)=\int_0^x f(t) dt$, then by definition $F(0)=0$, so the expression there can be written as $\frac{F(x)-F(0)}{x-0}$. So the limit is again by definition $F'(0)$.
– Ian
Commented Jan 30, 2015 at 1:55
• @Ian yes, this is clear. Commented Jan 30, 2015 at 1:56

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.

$\lim_{x \to 0}\frac{1}{x}\int_0^xf(t)dt = \lim_{x \to 0}\frac{\int_0^xf(t)dt-\int_0^0f(t)dt}{x-0}= \frac{d}{dx}\int_0^xf(t)dt |_{x=0} = f(x) |_{x=0} = f(0)$

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.