Indeterminate limit

  • #1
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##\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?
 

Answers and Replies

  • #2
mathman
Science Advisor
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461
Yes. I guess you are learning about Hospital's rule.
 
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