proof: $\lim_{n\rightarrow \infty } \int_{0}^{a}f(nx)dx = aL$ when $\lim{x\to\infty }f(x)= L$

Question

we know that:

$\lim{x\to\infty }f(x)= L$

$\alpha > 0$

i need to prove that:

$\lim_{n\rightarrow \infty } \int_{0}^{a}f(nx)dx = aL$

my idea was to get the limit inside the integral and then i will have this:

$\int_{0}^{a}L = L(a-0)$

but i am not sure that it is correct

Answer 1

Taking the limit inside the integral needs justification.

Let $\epsilon >0$ and choose $T$ such that $|f(x)-L| <\epsilon$ for $x >T$, Note that $\int_0^{a}f(nx)dx=\frac 1n \int_0^{na} f(y)dy=\frac 1n \int_0^{T} f(y)dy+\frac 1n \int_T^{na} f(y)dy$. Show that the second term has values between $\frac {na-T} n (L-\epsilon)$ and $\frac {na-T} n (L+\epsilon)$. Also the first term tends to $0$ as $n \to \infty$. Can you put these together to finish the proof?