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## Application of L’Hospital’s Rule

Solution: Note that $f(x)=\dfrac{e^xf(x)}{e^x}$, hence we can apply L’Hospital’s Rule
\begin{align*}
\lim_{x\to\infty}f(x)
=&\ \lim_{x\to\infty}\frac{e^xf(x)}{e^x}\\
$$\lim_{x\to\infty} f’(x)=\lim_{x\to\infty} (f(x)+f’(x))-\lim_{x\to\infty} f(x)=L-L=0.$$