# Silly Limit with Exponetiation

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

• How did you use L'Hôpital here? – Git Gud Jul 26 '13 at 20:50
• Take the logarithm, $\sqrt{x}\cdot\log x \to 0$ for $x \to 0^+$. So exponentiating again, we get $1$ for the limit. – Daniel Fischer Jul 26 '13 at 20:51

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 $\displaystyle\lim_{x\to 0^+} x^{\sqrt{x}}$.
• 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)... – colormegone Jul 26 '13 at 21:10
• I was referring to the $x^x$, in the guise $t^t$ where $t=\sqrt{x}$. – André Nicolas Jul 26 '13 at 21:14
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 $\exp(0)=1$.