Find $\lim\limits_{x \to \infty}{\mathrm{e}^{-x}\int_{0}^{x}{f\left(y\right)\mathrm{e}^{y}\,\mathrm{d}y}}$

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: $$\int_N^\infty f(x)e^xdx\geq\frac{1}{2}\int_N^\infty e^xdx$$

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}