Find $\lim\limits_{x \to \infty}{\mathrm{e}^{-x}\int_{0}^{x}{f\left(y\right)\mathrm{e}^{y}\,\mathrm{d}y}}$ Given $f(x)$ is a continuous function defined in $(0,\: \infty)$ such that $$\lim_{x \to \infty}f(x)=1$$ Then Find $$L=\lim_{x \to \infty}{\mathrm{e}^{-x}\int_{0}^{x}{f\left(y\right)\mathrm{e}^{y}\,\mathrm{d}y}}$$ 
My try:
we have $$L=\lim_{x \to \infty}{\frac{\int_{0}^{x}{f\left(y\right)\mathrm{e}^{y}\,\mathrm{d}y}}{e^x}}$$
If the numerator is finite then $L=0$ else by L'Hopital's Rule we have $\infty/\infty$ form we get
$$L=\lim_{x \to \infty}\frac{f\left(x\right)\mathrm{e}^{x}}{\mathrm{e}^x}=\lim_{x \to \infty}{f\left(x\right)}=1$$
But how to tell whether $\lim\limits_{x \to \infty}\int_{0}^{x}{f\left(y\right)\mathrm{e}^{y}\,\mathrm{d}y}$ is Finite or Infinite?
 A: Hint: Since $\lim_{x\to\infty}f(x)=1$, we have $f(x)>1/2$ for all $x>N$, where $N$ is sufficiently large. Ignoring the integral over $[0,N]$ (which is a fixed finite value), we consider the remaining part:
$$\int_N^\infty f(x)e^xdx\geq\frac{1}{2}\int_N^\infty e^xdx$$
A: Let $ \varepsilon>0 $, since $ \lim\limits_{x\to +\infty}{f\left(x\right)}=1 $, there exists some $ x_{0}>0 $ such that $ \left(\forall x\geq x_{0}\right),\ \left|f\left(x\right)-1\right|<\varepsilon \cdot $
Let $ x\geq x_{0} $, we have : \begin{aligned}\left|\mathrm{e}^{-x}\int_{0}^{x}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}\right|&=\left|\mathrm{e}^{-x}\int_{0}^{x_{0}}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}+\mathrm{e}^{-x}\int_{x_{0}}^{x}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}\right|\\ &\leq\mathrm{e}^{-x}\int_{0}^{x_{0}}{\mathrm{e}^{y}\left|f\left(y\right)-1\right|\mathrm{d}y}+\mathrm{e}^{-x}\int_{x_{0}}^{x}{\mathrm{e}^{y}\left|f\left(y\right)-1\right|\mathrm{d}y}\\ &\leq\mathrm{e}^{-x}\int_{0}^{x_{0}}{\mathrm{e}^{y}\left|f\left(y\right)-1\right|\mathrm{d}y}+\varepsilon\,\mathrm{e}^{-x}\int_{x_{0}}^{x}{\mathrm{e}^{y}\,\mathrm{d}y}\end{aligned}
$ \mathrm{e}^{-x}\int_{0}^{x_{0}}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}\underset{x\to +\infty}{\longrightarrow}0 $, thus, there exists some $ x_{1}>0 $ such that $ \mathrm{e}^{-x}\int_{0}^{x_{0}}{\mathrm{e}^{y}\left|f\left(y\right)-1\right|\mathrm{d}y}<\varepsilon $, hence, if $ x\geq \max{\left(x_{0},x_{1}\right)} $, we have : \begin{aligned}\left|\mathrm{e}^{-x}\int_{0}^{x}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}\right|\leq \varepsilon+\varepsilon\,\mathrm{e}^{-x}\int_{x_{0}}^{x}{\mathrm{e}^{y}\,\mathrm{d}y}\leq\varepsilon+\varepsilon\,\mathrm{e}^{-x}\int_{-\infty}^{x}{\mathrm{e}^{y}\,\mathrm{d}y}=2\varepsilon\end{aligned}
Thus $$ \lim_{x\to +\infty}{\mathrm{e}^{-x}\int_{0}^{x}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}}=0 $$
And we have : \begin{aligned} \mathrm{e}^{-x}\int_{0}^{x}{\mathrm{e}^{y}f\left(y\right)\mathrm{d}y}&=\mathrm{e}^{-x}\int_{0}^{x}{\mathrm{e}^{y}\,\mathrm{d}y}+\mathrm{e}^{-x}\int_{0}^{x}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}\\&=1-\mathrm{e}^{-x}+\mathrm{e}^{-x}\int_{0}^{x}{\mathrm{e}^{y}\left(f\left(y\right)-1\right)\mathrm{d}y}\underset{x\to +\infty}{\longrightarrow}1\end{aligned}
A: As long as the denominator tends to $\infty$ we can always apply L'Hopital's Rule. So the answer is $1$. (You need not bother about the numerator being bounded; it will never be). 
