- #1

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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?

- I
- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

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?

- #2

mathman

Science Advisor

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

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