Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 74

Evaluate the $\displaystyle \lim_{x \to \infty} \frac{e^{4x} - 1 - 4x}{x^2}$

By applying L' Hospitals Rule..


$
\begin{equation}
\begin{aligned}

\lim_{x \to \infty} \frac{e^{4x} - 1 - 4x}{x^2} =& \lim_{x \to \infty} \frac{e^{4x} (4) - 0 - 4(1)}{2x} = \lim_{x \to \infty} \frac{4(e^{4x} - 1)}{2x}


\end{aligned}
\end{equation}
$


If we evaluate the unit, we will still get an indeterminate form, so we must apply the L' Hospitals Rule once more. Thus,


$
\begin{equation}
\begin{aligned}

\lim_{x \to \infty} \frac{4(e^{4x} - 1)}{2x} = \lim_{x \to \infty} \frac{4(e^{4x})(4)}{2} =& \lim_{x \to \infty} \frac{16 e^{4x}}{2}
\\
\\
=& \frac{16 e^{4 \infty}}{2}
\\
\\
=& \infty

\end{aligned}
\end{equation}
$