How to find $\lim_{x \to 0^+} x^{\sqrt{x}}$ using L'Hospital's Rule.
Matthew Martinez 
I am trying to find $\lim_{x \to 0^+} x^{\sqrt{x}}$ using L'Hospital's Rule. I've differentiate the function, but it doesn't seem like that has helped at all.
$$\lim_{x \to 0^+} x^{\sqrt{x}} = \lim_{x \to 0^+} x^{\sqrt{x}} (\frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}})$$
$$\frac{d}{dx} y = \frac{d}{dx} \ln(x^{\sqrt{x}}) \to \frac{1}{y} \cdot y'= \sqrt{x}\ln(x) \to y' = y(\frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}}) \to y' = x^{\sqrt{x}} (\frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}})$$
How can I figure out what this limit represents?
$\endgroup$ 54 Answers
$\begingroup$Note first that$$\lim_{x\to 0+} x^x = \exp\left(\lim_{x\to 0^+} x \log x\right) = \exp\left(\lim_{x\to 0^+} \frac{\log x}{x^{-1}}\right) = \exp\left(-\lim_{x\to 0^+} x\right) = 1. $$by (continuity and) l'Hopital's rule. Thus$$\lim_{x\to 0+} x^{\sqrt{x}} = \lim_{x\to 0^+} x^{2x} = 1.$$
$\endgroup$ $\begingroup$Let $f(x) = x^\sqrt x$. Then $\ln f(x) = \sqrt x \ \ln(x) = \dfrac{\ln x}{x^{-\frac 12}}$.
Then $\dfrac{(\ln x)'}{\left( \dfrac{1}{\sqrt x} \right)'} =\dfrac{x^{-1}}{\frac 12 x^{-\frac 32}} = \sqrt x \to 0 \ \text{as} \ x \to 0^+$.
It follows that $f(x) \to e^0 = 1$ as $x \to 0^+$.
$\endgroup$ $\begingroup$You can transform it to a well-known limit:$$\lim_{x \to 0^+} x^{\sqrt{x}}=\lim_{x \to 0^+} [(\sqrt{x})^2]^{\sqrt{x}}=\lim_{x \to 0^+} [(\sqrt{x})^{\sqrt{x}}]^2=\lim_{t \to 0^+} [t^t]^2=\left[\lim_{t \to 0^+} t^t\right]^2=1^2=1.$$For $\lim_\limits{t \to 0^+} t^t=1$, refer here or here.
$\endgroup$ $\begingroup$$$ \lim_{x\to 0^+}\ln{x^{\sqrt{x}}}= \lim_{x\to 0^+}\sqrt{x}\ln{x}= \lim_{x\to 0^+}\frac{\ln{x}}{\frac{1}{\sqrt{x}}}\stackrel{\text{L'H}}{=}\\ \lim_{x\to 0^+}\frac{\left(\ln{x}\right)'}{\left(\frac{1}{\sqrt{x}}\right)'}= \lim_{x\to 0^+}\frac{\frac{1}{x}}{-\frac{1}{2\sqrt{x^3}}}= -2\lim_{x\to 0^+}\sqrt{x}=-2\cdot 0=0.\\ \ln{x^{\sqrt{x}}}\xrightarrow{x\rightarrow 0^+}0\implies x^{\sqrt{x}}\xrightarrow{x\rightarrow 0^+}1. $$Therefore:
$$ \lim_{x\to 0^+}x^{\sqrt{x}}=1. $$
I used the simple fact that if the logarithm of a function of x ($f(x)=x^{\sqrt{x}}$) approaches 0 as x goes to 0 from the right ($\lim_{x\to 0^+}\ln{f(x)}=0$), $f(x)$ must approach $1$ because only the logarithm of a quantity that's getting closer to 1 is getting closer to 0.
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