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,


$\begin{equation} \begin{aligned} \lim_{x \to \infty} \frac{4(e^{4x} - 1)}{2x} = \lim_{x \to \infty} \frac{4(e^{4x})(4)}{2} =& \lim_{x \to \infty} \frac{16 e^{4x}}{2} \\ \\ =& \frac{16 e^{4 \infty}}{2} \\ \\ =& \infty \end{aligned} \end{equation}$