question about improper integrals and limits

Question

Let be $f:[0,\infty[ \rightarrow \mathbb{R}$ continuous. If $\displaystyle \int_0^{\infty} f(x) dx $ converges so $\displaystyle \lim_{r\rightarrow \infty} \int_r^{\infty} f(x) dx=0$

My Attempt:

Call $G(r) = \displaystyle \int_0^{r} f(x)dx, \hspace{0.2cm}$ then $ \hspace{0.2cm}\displaystyle \lim_{r\rightarrow \infty} G(r) = L$

and we have $L = \displaystyle \lim_{r\rightarrow \infty} G(r) = \displaystyle \lim_{r\rightarrow \infty} \displaystyle \int_0^{\infty} f(x) dx = \lim_{r\rightarrow \infty} \displaystyle \int_0^{r} f(x) dx$ + $\displaystyle \int_r^{\infty} f(x) dx = \lim_{r\rightarrow \infty} \displaystyle \int_0^{r} f(x) dx + \lim_{r\rightarrow \infty} \displaystyle \int_r^{\infty} f(x) dx = L + \lim_{r\rightarrow \infty} \displaystyle \int_r^{\infty} f(x) dx $

Then $\displaystyle \lim_{r\rightarrow \infty} \displaystyle \int_r^{\infty} f(x) dx= L-L = 0$

But I think that this isn't correct because this expression $ \displaystyle\lim_{r\rightarrow \infty} \displaystyle \int_0^{r} f(x) dx + \lim_{r\rightarrow \infty} \displaystyle \int_r^{\infty} f(x) dx $,

Is there a way to change this expression or another way to solve this exercise?

Answer 1

I think it is safer this way: You know that $\lim_{r\to\infty}\int_0^rf\,dx=L$. Choose $R$ large enough such that for any $R<r_1<r_2$, $|\int_0^{r_1}f\,dx-L|<\epsilon/2$ and $|\int_0^{r_2}f\,dx-L|<\epsilon/2$. Then $|\int_{r_1}^{r_2}f\,dx|<\epsilon$ and, a fortiori, $|\int_{r_1}^{\infty}f\,dx|<\epsilon$ (it is easy to see that the integral $\int_{r_1}^{\infty}f\,dx$ is convergent). By definition this means that $\lim_{r\to\infty}\int_{r}^{\infty}f\,dx=0$.