TITLE: method to find Limit of integral
QUESTION [6 upvotes]: In a multiple choice question, the following is asked :
The limit $$\lim\limits_{x \to 0^+} \displaystyle\int_x^1 \left(1+ \dfrac 1t\right) e^{-\tfrac 1t} dt $$
is equal to :

$+\infty$
$-\infty$
$e^{-1}$
0
1

My choice went for (3) because the integral is convergent and the integrand is positive and - unless mistaken - it is bounded between $0$ and $\dfrac 2e$.
I would like to know whether what I did is correct and hopefully how to prove that the value of the limit is indeed $\dfrac 1e$

REPLY [4 votes]: By a change of variable, the integral is
$$\int_1^\infty \frac{1+u}{u^2}e^{-u}du.$$
As
$$0<\frac{1+u}{u^2}e^{-u}< e^{1-u},$$
the integral is smaller than $1$, so 3.

Also note that the integrand is the derivative of
$$-\frac{e^{-u}}u.$$