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  1. $\displaystyle\lim_{x\to\frac{\pi}{2}}(\sin x)^{\tan x}$

  2. $\displaystyle\lim_{x\to0}x^2\ln x$

  3. $\displaystyle\lim_{x\to1^+}x^{\frac{1}{1-x}}$

Do I have to apply l'Hôpital's Rule to evaluate these limits?

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    $\begingroup$ Could you tell us your opinion ? $\endgroup$ – Claude Leibovici Feb 22 '14 at 9:27
  • $\begingroup$ It provides the simplest path for the first two, and works for the third. But the third can be done faster without. $\endgroup$ – André Nicolas Feb 22 '14 at 9:27
  • $\begingroup$ Please use MathJax in future $\ddot\smile$ $\endgroup$ – Shaun Feb 22 '14 at 9:41
  • $\begingroup$ Careful! It may help to get the limit into the correct form for l'Hopital's Rule, but it might not always be applicable. I'm ashamed to admit I'm not sure if l'Hopital's Rule could be used for the third, but I think I see another way of doing it. $\endgroup$ – Mike Feb 22 '14 at 10:51
  • $\begingroup$ Oops. Forget what I said on number 3. For some reason, I was thinking the negative part would be inside the logarithm... $\endgroup$ – Mike Feb 22 '14 at 11:43
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For the second limit: Yes $$\lim_{x\to 0}x^2\log x=\lim_{x\to 0}\frac{\log x}{\frac1{x^2}}=\lim_{x\to 0}\frac{\frac 1x}{-2x^{-3}}=-\lim_{x\to 0}\frac 12x^2=0$$ and for the other limits it's more simple to use the Taylor series.

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In 1. and 3., take the logarithm of the function and then write it as a fraction.

In 2., $x^2\ln x=\dfrac{x^2}{\frac{1}{\ln x}}$.

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