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

Question

Evaluate $$\lim \limits_{x \to \infty} \frac{1}{e^{x^2}}\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt.$$

I should probably use such inequality:

$$ \int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt > \frac{e^{x^2} \ln x}{2x} \text { for } x \ge 1,$$

which gives me that the numerator goes to infinity. I used L'Hopital rule, but without any conclusions. $$\frac{d}{dx}(\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt) = e^{x^2} (-1 + e^{\frac{\ln^2(x)}{x^2}} (1 + x^2) - e^{\frac{\ln^2(x)}{x^2}} \ln(x))$$ and $$\frac{d}{dx}(e^{x^2}) = 2 x e^{x^2}.$$

Please help.

Answer 1

Let $$I(x) =\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt \Rightarrow I(x) \geq \frac{\ln x}{x}e^{x^2} \stackrel{x\rightarrow\infty}{\longrightarrow}\infty$$ So, we have with $\frac{I(x)}{e^{x^2}}$ a L'Hospital case of $\frac{\infty}{\infty}$ for $x\rightarrow\infty$: $$\frac{I(x)}{e^{x^2}} \sim \frac{e^{(x^2+2\ln x+ \frac{\ln^2x}{x^2})}(\frac{x^2-\ln x + 1}{x^2}) - e^{x^2}}{2xe^{x^2}}=\frac{e^{\frac{\ln^2x}{x^2}}(x^2-\ln x +1)}{2x}- \frac{1}{2x} = \frac{1}{2}e^{\frac{\ln^2x}{x^2}} \left(x - \frac{\ln x}{x} + \frac{1}{x} \right)- \frac{1}{2x}\stackrel{x\rightarrow\infty}{\longrightarrow}\infty$$

Answer 2

You are on the right track. By using L'Hopital you have to consider the ratio of the two derivatives: $$\frac{ -1 + e^{\frac{\ln^2(x)}{x^2}} (1 + x^2 -\ln(x))}{ 2 x}= \frac{ -1 + (1+\frac{\ln^2(x)}{x^2}+o(\frac{\ln^2(x)}{x^2})) (1 + x^2 -\ln(x))}{ 2 x}\\ =\frac{x^2+O(\ln^2(x)))}{ 2 x}=\frac{x+O(\ln^2(x)/x))}{ 2}\to +\infty$$ where we used the fact that as $x\to +\infty$, $\ln(x)/x\to 0$ and $e^t=1+t+o(t)$ for $t\to 0$.