Evaluate $ \lim_{x \to \infty} e^{-x^2} \int_x^{x + \ln(x)/x} e^{-t^2} dt $

I've been trying all day to solve this and I can't figure it out how to do it, I know you have to use L'Hopital and that is a similar one to another limit that is already posted here, but the exponential having a minus complicates everything.

I need help, please.

Solution 1:

Let's start by simplifying the expression.

First, we'll use a limit law, saying the limit of a product is a product of the limits.

$$\lim_{x\to \infty} e^{-x^2} \cdot \lim_{x\to\infty} \int^{x+\frac{\ln(x)}{x}}_x e^{-t^2}dt$$

Let's evaluate the first limit.

$$\lim_{x\to\infty}e^{-x^2}=\lim_{x\to\infty}\frac{1}{e^{x^2}}$$

As $x$ approaches $\infty$, $e^{x^2}$ also approaches $\infty$, therefore $\frac{1}{e^{x^2}}$ approaches $0$, therefore one of the limits calculated is $0$.

However, we're not done. If the limit of the integral is $\infty$, we would get an indeterminate form. By monotonicity, $$|{\int_x^{x+\frac{\ln(x)}{x}} e^{-t^2}dt}|\leq (x+\frac{\ln(x)}{x}-x)e^{-x^2}=\frac{\ln(x)}{x}e^{-x^2}$$, which if we evaluate as $x$ approaches $\infty$, we get $0 \cdot 0 =0$, so we didn't have to use L'Hôpital's rule.

Hope this helps!

P.S. Special thanks to Gary for the monotoncity hints!