Evaluate the limit $\lim_{t\rightarrow\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)$ $$\lim_{t\rightarrow\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)$$
I have no idea where to start. Any help will be appreciated!
 A: Rewrite the limit as a $\frac{0}{0}$ indeterminate form and use L'Hopital.
$$\begin{align}
\lim_{t\rightarrow\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)&=\lim_{t\rightarrow\infty}\frac{\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s}{e^{-t}/t}\\
&=\lim_{t\rightarrow\infty}\frac{-e^{-t}/t}{-\left(\frac{e^{-t}(t+1)}{t^2}\right)}\\
&=\lim_{t\rightarrow\infty}\frac{t}{t+1}
\end{align}$$
A: We can use the Incomplete Gamma function just for a little fun.
$$ \Gamma(a,z)=\int_z^{\infty}t^{a-1}e^{-t}dt   $$
$$\lim_{t\to\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}ds\right)= \lim_{t\to\infty}\left(te^t\Gamma(0,t)\right) $$
Now the series expansion of the limit...
$$\lim_{t\to\infty}\left[ te^te^{-t}\left(\frac{1}{t}-\frac{1}{t^2}+\frac{2}{t^3}-\frac{6}{t^4}+\mathcal{O}\left(\frac{1}{t^5}\right)\right)\right]  $$
$$\implies \lim_{t\to\infty}\left[ 1-\frac{1}{t}+\frac{2}{t^2}-\frac{6}{t^3}+\mathcal{O}\left(\frac{1}{t^4}\right) \right]=\dots $$
A: Integrate by parts :$$te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s=1-te^t\int_t^\infty\frac{e^{-s}}{s^2}ds$$
It remains to see that $$\displaystyle \int_t^\infty\frac{e^{-s}}{s^2}ds = o\left(\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)$$

Lemma
Let $f: [a,\infty(\to \mathbb R$ and $\phi :  [a,\infty(\to \mathbb R^+$ be such that $f=o_\infty(\phi)$
If $\int_a^\infty\phi$ converges then $\int_x^\infty f(t)dt = o\left(\int_x^\infty \phi(t) dt\right)$

Using the lemma here yields $$te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s +o\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)=1$$
Hence the limit is $1$.
