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I need to compute $$\lim_{x\to 0^+}x^\epsilon \ln x, \epsilon>0$$

I tried to represent the limit as $$\dfrac{x^{\epsilon +1}\ln x}{x}$$

but it didn't work. Can you hint me?

john
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2 Answers2

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Using l'Hôpital rule:

$\lim_{x \rightarrow 0^{+}} x^{\epsilon}ln(x) = \lim_{x \rightarrow 0^{+}}\frac{ln(x)}{x^{-\epsilon}} = \lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\epsilon x^{-\epsilon-1} } = \lim_{x \rightarrow 0^{+}} \frac{-x^{\epsilon}}{\epsilon} = 0.$

Alex Silva
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As you did not mention in the OP that you want to you L'Hopital, what if I told you L'Hopital is not needed here? If you put $y = \frac1{x^\epsilon}$ you get a limit $-\frac1\epsilon\lim_{y\to+\infty}\frac{\log y}{y} =0.$

SBF
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  • I understand you! But it's my exercise to solve it using L'Hôpital's rule. Anyway thanks for your help! :) – john Dec 05 '14 at 11:05