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I'm trying to solve the expression $$\lim_{x\to\infty} \frac{x^{2x}}{x^{5x}}.$$ I know the limit is equal to $0,$ and since the numerator and denominator go to infinity as $x$ goes to infinity, I should be able to use L'Hôpital's rule but it doesn't seem to work.

Is this the wrong approach? If so, how would you solve this?

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    $\begingroup$ Rewrite using logarithm $\endgroup$
    – j. kookalinski
    Feb 26, 2018 at 18:34
  • $\begingroup$ What is your question? You ask one question in the title, and another, different expression in the body of the post. Please edit the limit in the body to match the limit in the title, or vice-versa, edit the limit in the title to match the limit in the body of your question. $\endgroup$
    – amWhy
    Feb 26, 2018 at 18:53
  • $\begingroup$ @TonyS.F. I think there's a much simpler approach, based only on the laws of exponents. $\endgroup$
    – amWhy
    Feb 26, 2018 at 18:55
  • $\begingroup$ @amWhy you're right, see the answer by vadim123 $\endgroup$
    – j. kookalinski
    Feb 26, 2018 at 18:56

2 Answers 2

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$$\frac{x^{2x}}{x^{5x}}=x^{-3x}=\frac{1}{(x^x)^3}$$ No need for L'Hopital's rule, as the denominator goes to $\infty $ as $x\to\infty$.

The question in the title is different, but solved the same way.

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$$\frac {x^{2x}}{x^{5x}}=x^{2x-5x}=x^{-3x} $$

$$=e^{-3x\ln (x)} $$

When $x\to +\infty, -3x\ln (x)\to-\infty $ and

$$e^{-3x\ln (x)}\to 0$$

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