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I found, using exponetiation and l'hopital, that:

$\lim \limits_{x\to 0} x^{\sqrt x} = \infty$

However, the limit is 1 using matlab. How can I find such limit?

Thanks

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    $\begingroup$ How did you use L'Hôpital here? $\endgroup$ – Git Gud Jul 26 '13 at 20:50
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    $\begingroup$ Take the logarithm, $\sqrt{x}\cdot\log x \to 0$ for $x \to 0^+$. So exponentiating again, we get $1$ for the limit. $\endgroup$ – Daniel Fischer Jul 26 '13 at 20:51
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Rewrite as $(\sqrt{x}^{\sqrt{x}})(\sqrt{x}^{\sqrt{x}})$ and quote the presumably familiar result that $\lim_{y\to 0^+}y^y=1$. So the answer is $(1)(1)$.

Note that the question should really ask for $\displaystyle\lim_{x\to 0^+} x^{\sqrt{x}}$.

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  • $\begingroup$ One might also use $ \ \lim_{x \rightarrow 0^{+}} (x^{x^{(1/2)}}) = \lim_{x \rightarrow 0^{+}} (x^{x/2}) = (\lim_{x \rightarrow 0^{+}} x^x)^{(1/2)} , \ $ since $ \ \lim_{x \rightarrow 0^{+}} x^x \ $ might be a more familiar result (that is, one the instructor or the book did, or that the student has already been asked to establish)... $\endgroup$ – colormegone Jul 26 '13 at 21:10
  • $\begingroup$ I was referring to the $x^x$, in the guise $t^t$ where $t=\sqrt{x}$. $\endgroup$ – André Nicolas Jul 26 '13 at 21:14
  • $\begingroup$ All right, I was just wondering whether OP would understand the slight subtlety there... $\endgroup$ – colormegone Jul 26 '13 at 21:17
  • $\begingroup$ Thank you for your comment, it helped me add a sentence to the answer. $\endgroup$ – André Nicolas Jul 26 '13 at 21:22
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You can make the substitution $\sqrt{x}=t$ and compute $$ \lim_{t\to0^+}(t^2)^t=\lim_{t\to0^+}t^{2t}= \lim_{t\to0^+}\exp(2t\log t) $$ Since $\lim_{t\to0^+}t\log t=0$ (it's a standard exercise), you can conclude that your limit is $\exp(0)=1$.

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