Find this limit

Relevant wiki: L'Hôpital's Rule .

$\displaystyle \lim_{x \to \infty} \dfrac{x^2}{e^{4x}-1-4x}$

$\displaystyle = \lim_{x \to \infty} \dfrac{2x}{4e^{4x} -4}$

$\displaystyle = \lim_{x \to \infty} \dfrac{x}{2(e^{4x} -1)}$

$\displaystyle = \lim_{x \to \infty} \dfrac{1}{8e^{4x}}$

$\displaystyle = \dfrac 18 \times \lim_{x \to \infty} \dfrac{1}{e^{4x}}$

$= \dfrac 18 \times \dfrac 1{\infty}$

$= \boxed{0}$

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{x \to \infty} \frac {x^2}{{\color{#3D99F6}e^{4x}}-1-4x} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to \infty} \frac {x^2}{{\color{#3D99F6}\left(1+4x+\frac {4^2x^2}{2!} + \frac {4^3x^3}{3!}+ \cdots \right)}-1-4x} \\ & = \lim_{x \to \infty} \frac {x^2}{\frac {4^2x^2}{2!} + \frac {4^3x^3}{3!} + \frac {4^4x^4}{4!} \cdots} & \small \color{#3D99F6} \text{Divide up and down by }x^2 \\ & = \lim_{x \to \infty} \frac 1{\frac {4^2}{2!} + \frac {4^3x}{3!} + \frac {4^4x^2}{4!} \cdots} \\ & = \boxed{0} \end{aligned}$