Find $\lim_{x\to 0^+} \int_x^{+\infty} t^{-1}e^{-t}dt/{\ln(1/x)}$ Find $$\lim_{x\to 0^+}\frac{\int_x^{+\infty} t^{-1}e^{-t}dt}{\ln\frac1x}$$
I found this problem and I need some help understanding it
In the solution, it said this
First the integral $$\int_a^{\infty} t^{-1}e^{-t}dt$$ is convergent for every fixed number $a \gt0$ and $\ln\frac1x \to +\infty$ when $x \to +0$
And this is what I tried to show that $\int_a^{\infty} t^{-1}e^{-t}dt$ is convergent.I used the first mean value theorem, therefore there exist $c\in[a,+\infty)$ such that 
$$\int_a^{\infty} t^{-1}e^{-t}dt=\frac1c\int_a^{\infty}e^{-t}dt=\frac1{e^ac}$$ 
since $\frac1{e^ac}$ is a finite number this means that the integral converges (I hope this proof is correct)
this implies that $$\lim_{x\to 0^+}\frac{\int_a^{+\infty} t^{-1}e^{-t}dt}{\ln\frac1x}=0$$
the solution continues and it says
By using L'Hopital rule
$$\lim_{x\to 0^+}\frac{\int_x^{+\infty} t^{-1}e^{-t}dt}{\ln\frac1x}=\lim_{x\to 0^+}\frac{\int_x^{a} t^{-1}e^{-t}dt}{\ln\frac1x}=\lim_{x\to +0}\frac{-x^{-1}e^{-x}}{-x^{-1}}=\lim_{x\to +0}e^{-x}=1$$ 
But the part that I don't understand in this is how did they go from $\int_x^{+\infty} t^{-1}e^{-t}dt$ to $\int_x^{a} t^{-1}e^{-t}dt $ and how did they now that that the integral $\int_x^{a} t^{-1}e^{-t}dt $  tends to infinity because to use L'Hopital rule that must tend to infinity
 A: Hint. Assume $0<x<a$. One may see that
$$
t^{-1}e^{-a} \le t^{-1}e^{-t} \le t^{-1}e^{-x}
$$ giving (by integrating)
$$
e^{-a}\left(\ln a - \ln x \right)\le \int_x^{a} t^{-1}e^{-t}dt \le e^{-x}\left(\ln a - \ln x \right)
$$ then, by the squeeze theorem, the integral tends to $\infty$ as $x \to 0^+$.
One may also notice that
$$
\int_x^{\infty} t^{-1}e^{-t}dt=\int_x^{a} t^{-1}e^{-t}dt+\int_a^{\infty} t^{-1}e^{-t}dt
$$ and the latter integral is convergent.
A: One way to understand the integral $\int_x^{+\infty} t^{-1}e^{-t}dt$ is
to write it in the form
$$\int_x^{1} {1 \over t} dt + \int_x^{1} {e^{-t} - 1 \over t}dt + 
\int_{1}^{\infty} {e^{-t} \over t}$$
Because ${\displaystyle {e^{-t} - 1 \over t}}$ converges to $1$ as $x$ goes to $0$, the
second integral converges as $x$ goes to zero. This shows divergence 
of the overall limit as $x$ goes to zero since the first term is ${\displaystyle \ln {1 \over x}}$. And it's not hard to go from here to show that your original limit is just $1$.
A: $$lim_{x\to 0}\frac{\int_x^{\infty}t^{-1}e^{-t}dt}{\ln\left(\frac{1}{x}\right)}=lim_{x\to 0}\left(-\frac{\int_x^{\infty}t^{-1}e^{-t}dt}{\ln(x)}\right)=lim_{x\to 0}\left(-\frac{-(x^{-1}e^{-x})+\int_x^{\infty}0\,dt}{\frac{1}{x}}\right)=\lim_{x\to 0}\left(e^{-x}\right)=1$$
