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 :

1. $+\infty$
2. $-\infty$
3. $e^{-1}$
4. 0
5. 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 $\dfrac 1e$