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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}\\
\text{Apply L’Hospital’s Rule}\quad
=&\ \lim_{x\to\infty}\frac{e^xf(x)+e^xf’(x)}{e^x}\\
=&\ \lim_{x\to\infty}(f(x)+f’(x))\\
=&\ L.
\end{align*} Therefore, we also have
$$
\lim_{x\to\infty} f’(x)=\lim_{x\to\infty} (f(x)+f’(x))-\lim_{x\to\infty} f(x)=L-L=0.
$$


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