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