A continuous function $f:[0,\infty)\to \mathbb{R}$ such that $\lim\limits_{x\to \infty} \left(f(x)+\int_0^x f(t) dt \right)=0$

Question

Let $f:[0,\infty)\to \mathbb{R}$ be a continuous function such that $\lim\limits_{x\to \infty} \left(f(x)+\int_0^x f(t) dt \right)=0$. Prove that $$\lim \limits_{x\to \infty} \int_0^x f(t)dt=0.$$
I don't understand the solution presented in my book. They start by saying that $$\lim_{x\to\infty}\int_0^x f(t) dt= \lim_{x\to \infty} \frac{e^x \cdot \int_0^x f(t)dt}{e^x}$$and then they apply L'Hospital's rule. Why is this allowed? We don't know the limit of the numerator, so we can't be sure that this is $\frac{\infty}{\infty}$.

Answer 1

I'll try to prove is directly, avoiding L'Hopital.

Let $g(x) =f(x)+\int_0^x f(t) dt =F'(x)+F(x) $ where $F(x) =\int_0^x f(t) dt $.

Then $g(x) \to 0$ as $x \to \infty$.

Then $(e^xF(x))' =e^x(F'(x)+F(x)) =e^xg(x) $ so $$(e^xF(x))|_a^b =\int_a^b e^xF(x) dx =\int_a^b e^xg(x)dx $$

For any $c > 0$, choose $a$ such that $|g(x)| < c$ for $x \ge a$.

Then, for any $b > a$,

$\begin{array}\\ |\int_a^b e^xg(x)dx| &\le c|\int_a^b e^xdx|\\ &=c(e^b-e^a)\\ &\le ce^b\\ \text{and}\\ |(e^xF(x))|_a^b| &=|e^bF(b)-e^aF(a)|\\ &\ge|e^bF(b)|-|e^aF(a)|\\ \text{so}\\ |e^bF(b)|-|e^aF(a)| &\le ce^b\\ \text{or}\\ |e^bF(b)| &\le ce^b+|e^aF(a)|\\ \text{or}\\ |F(b)| &\le c+e^{-b}|e^aF(a)|\\ &\le 2c \qquad\text{by choosing }e^{b}>|e^aF(a)|/c\\ \end{array} $

Therefore $|F(b)| < 2c$ for all $b > \max(a, \ln(|e^aF(a)|/c)) $ where $|g(x)| < c$ for $x > a$, so that $F(x) \to 0$.