Evaluate the limit using l'Hôpital's Rule. $\lim _{x \rightarrow \infty} \frac{(\ln x)^{3}}{4 x^{2}}$
A) 3
B) $\frac{3}{8}$
C) $\frac{1}{4}$
D) 0

Question

Quizplus · Accepted Answer

Applying L'Hopital's rule, we get:

\begin{align*}
\lim_{x	o\infty}\frac{(\ln x)^3}{4x^2}&\stackrel{\frac{\infty}{\infty}}{=}\lim_{x	o\infty}\frac{3(\ln x)^2\cdot\frac{1}{x}}{8x}\
&\stackrel{\frac{\infty}{\infty}}{=}\lim_{x	o\infty}\frac{3\ln x\cdot\frac{1}{x}}{8}\
&\stackrel{\frac{0}{\infty}}{=}0
\end{align*}

Therefore, the limit is 0.

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Evaluate the Limit Using L'Hôpital's Rule $\lim _{x \rightarrow \infty} \frac{(\ln x)^{3}}{4 x^{2}}$