First order approximation of $F(x)=\int_0^x f(t) dt$ in the neighbourhood of $\infty$

Question

Let $f(x)$ continuous on the real line. Then the first order approximation of $$F(x)=\int_0^x f(t) dt$$ in the neighbourhood of $0$ is: $$F(x)=\int_0^x f(t) dt\sim 0 + x f(0), \ \ \ (x\rightarrow 0)$$ But what is the first order approximation of $F(x)$ in the neighbourhood of $+\infty$?

Answer 1

Make the change of variable y=1/x. The integral is from o to 1/x while the integrand is f(1/x).(1/x^2). So the required value is (1/x).lim(f(1/x)/x^2), the limit being taken near 0 and we need this to exist.