Calculate the limit $\lim_{N \to \infty} e^N \int_{N}^{\infty} \frac{e^{-x}}{\log(x)}dx $ Does the following limit converges to $0$ or not? $$ \lim_{N\to \infty} e^N \int_{N}^{\infty} \frac{e^{-x}}{\log(x)} dx$$ I am interested whether the limit is $0$ or not. How can I get the answer? 
I have tried this: $\int_{N}^{\infty} \frac{e^{-x}}{\log(x)} < \frac{1}{\log(N)} \int_{N}^{\infty}e^{-x} = \frac{e^{-N}}{\log(N)}$ hence $$I_N < \frac{1}{\log(N)}$$ therefore $I_N \to 0$ for $N \to \infty$. Is it correct?
 A: HINT:
Rephrase the problem to get
$$\lim_{N\to\infty} \frac{\int_N^\infty \frac{e^{-x}}{\ln x}dx}{e^{-N}}$$
Now we have a fraction whose limits on the numerator and denominator are both zero, so perhaps you can use L'Hopital's rule to evaluate it.
Stop reading now if you want to try it yourself.
FULL ANSWER:
Let
$$f(x)=\frac{e^{-x}}{\ln x}$$
and let $F(x)$ be its antiderivative. Then we have
$$\lim_{N\to\infty} \frac{\lim_{k\to\infty} F(k)-F(N)}{e^{-N}}$$
To make a long story short, the integral converges, so $\lim_{k\to\infty} F(k)$ is a constant and when we take the derivative of the top and bottom, we get
$$\lim_{N\to\infty} \frac{-f(N)}{-e^{-N}}$$
$$\lim_{N\to\infty} \frac{\frac{e^{-N}}{\ln N}}{e^{-N}}$$
$$\lim_{N\to\infty} \frac{1}{\ln N}$$
Which is zero.
A: Your approach is correct. The wanted limit is
$$ \lim_{N\to +\infty}\int_{N}^{+\infty}\frac{e^{N-x}}{\log x}\,dx = \lim_{N\to +\infty}\int_{0}^{+\infty}\frac{e^{-u}}{\log(u+N)}\,du $$
and for any $N>1$ we have
$$ 0\leq \int_{0}^{+\infty}\frac{e^{-u}}{\log(u+N)}\,du \leq \frac{1}{\log N}\int_{0}^{+\infty}e^{-u}\,du = \frac{1}{\log N} $$
hence the limit is trivially $\color{red}{0}$ by squeezing. Indeed
$$ \int_{N}^{+\infty}\frac{dx}{e^x \log x} \sim \frac{1}{e^N \log N} $$
by the dominated convergence theorem.
