$\lim_{x\to 0^+}\left(x^\sqrt{x}\right)=?$

How do I turn it into a fraction? Is L'Hospital's rule even applicable?

  • $\begingroup$ You could try using logarithms $\endgroup$ – Triatticus Nov 10 '17 at 3:43

Hint: Consider $\log$, so $\sqrt{x}\log x=\dfrac{\log x}{\dfrac{1}{\sqrt{x}}}$.


given limit = $e^{\sqrt{x}\ln x} $

we know $\ln x < (x-1) < x $

$0 \leq\sqrt{x}\ln x < \sqrt{x} x $

this limit converges to 0 as $x\to 0$ by squeeze theorem

i.e., given limit converges to $e^0= 1$ as $x \to 0$


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