Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.6

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.

$$