Undergrad Indeterminate Limit: Evaluating ##\displaystyle \lim_{a \to 0^+} a^2 \log a##

[Mr Davis 97]

$\displaystyle \lim_{a \to 0^+} a^2 \log a = 0 \cdot (- \infty)$, which is an indeterminate form.

So $\displaystyle \lim_{a \to 0^+} a^2 \log a = \lim_{a \to 0^+} \frac{\log a}{a^{-2}} = \lim_{a \to 0^+} \frac{\frac{1}{a}}{(-2)a^{-3}} = -\frac{1}{2}\lim_{a \to 0^+} a^2 = 0$.

Is this correct?

 

[mathman]

Yes. I guess you are learning about Hospital's rule.