I was trying to preparing for Analysis exam and came across this question.
Suppose that $f: [0,1] \to \Bbb R$ is continuous. Suppose sequence $\{a_n\}_{n\ge1}>0$ with $$\lim_{n\to\infty}a_n = 0 $$ Prove that $$\lim_{n\to\infty} \ \frac {1}{a_n} \int_{0}^{a_n}f(x)\ dx = f(0)$$
I'm not quite sure what to do with it, but I tried the following: Exchange $a_n$ for a continuous and differentiable function $g(n)$ defined as $g: \Bbb R\to \Bbb R \;$ with $$\lim_{n\to\infty}g(n) = 0 $$ then the limit becomes $$\lim_{n\to\infty} \ \frac {1}{g(n)} \int_{0}^{g(n)}f(x)\ dx $$ and applying FTC $$\lim_{n\to\infty} \ \frac {1}{g(n)} \int_{0}^{g(n)}f(x)\ dx = \lim_{n\to\infty} \ \frac {F(g(n)) - F(0)}{g(n)} =$$ then using L'Hopital's rule $$= \lim_{n\to\infty} \ \frac {f(g(n))g'(n)}{g'(n)} = \lim_{n\to\infty} \ f(g(n)) = f(\lim_{n\to\infty}g(n)) = f(0)$$
which is the results that I needed to prove. However, I'm not sure how to show that same holds for a sequence $a_n$. Any hints?
Thank you.
${1\over{a_n}}\int_0^{a_n}f(x)dx={1\over{a_n}}a_nf(c_n)=f(c_n)$ where $c_n\in[0,a_n]$, we deduce that $lim_nc_n=0$ and $lim_n{1\over{a_n}}\int_0^{a_n}f(x)dx=lim_nf(c_n)=f(lim_nc_n)=f(0)$.