$\lim_{x \rightarrow 0}\frac{1}{x^2 \log x}$

$\lim_{x \rightarrow 0}\frac{1}{x^2 \log x} $

Calculus
Gubiv Gubiv
7
Report

1 Answer

We can use L'hopital's rule
\[\lim_{x \rightarrow 0}\frac{1}{x^2 \log x} =\lim_{x \rightarrow 0} \frac{1}{\frac{\log x}{\frac{1}{x^2}}}=\frac{1}{\lim_{x \rightarrow 0} \frac{\frac{1}{x}}{\frac{-2}{x^3}}}=\frac{1}{\lim_{x \rightarrow 0} -\frac{1}{2}x^2}=-\infty.\]

Erdos Erdos
4.8K
Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.