If $\int_a^{+\infty} f(x)dx $ converges, and $\lim_{x\to + \infty} f(x) = L$. Then prove that $L=0$

Question

I have the following statement which I have to prove.

If $\int_a^{+\infty} f(x)dx $ converges, and $\lim_{x\to + \infty} f(x) = L$. Then prove that $L=0$

The idea is to prove it only using the improper integral definition, but I don’t know how to do it. Thanks in advance!

Edit: I wrote down the limit definition for f, and then integrate both sides. Doing that I got: $\int_a^{+\infty} L-\varepsilon \ dx \leq L’ $ (Being $L’= \int_a^{+\infty} f(x)dx $ (Since it converges) That is the hint that the professor showed us in a drawing. But how can I use this information to conclude that L must be equal to 0?

Answer 1

Hint: Assume that $L \neq 0$, and prove that the integral diverges.

Answer 2

Note that for any $m\ge a$ we have $\lim_{n \to \infty} {1 \over n-m} \int_m^n f(x)dx =0$.

Let $\epsilon>0$ and choose $m \ge a$ such that $|f(x)-L| < \epsilon$. Then we see that $L-\epsilon \le \lim_{n \to \infty} {1 \over n-m} \int_m^n f(x)dx \le L+\epsilon$ and so $0 \in [L-\epsilon, L+\epsilon]$ for all $\epsilon>0$. Hence $L=0$.