Answer
$$\lim_{x\to0^+}x^{\sqrt x}=1$$
Work Step by Step
$$A=\lim_{x\to0^+}x^{\sqrt x}$$
We can write $x=e^{\ln x}$, which means we can also rewrite $x^{\sqrt x}$ into $(e^{\ln x})^{\sqrt x}$. That makes A into
$$A=\lim_{x\to0^+}(e^{\ln x})^{\sqrt x}$$
$$A=\lim_{x\to0^+}(e^{\sqrt x\ln x})$$
$$A=e^{\lim_{x\to0^+}(\sqrt x\ln x)}.$$
We now call $$B=\lim_{x\to0^+}\sqrt x\ln x$$
$$B=\lim_{x\to0^+}\frac{\sqrt x\ln x}{1}.$$
We now divide both numerator and denominator by $\sqrt x$, which means
$$B=\lim_{x\to0^+}\frac{\ln x}{\frac{1}{\sqrt x}}.$$
As $\lim_{x\to0^+}(\ln x)=\ln 0^+=-\infty$
and $\lim_{x\to0^+}\frac{1}{\sqrt x}=\frac{1}{\sqrt 0}=+\infty$ (only positive values of $x$ are concerned),
this is an indeterminate form of $-\infty/+\infty$. We will apply L'Hospital's Rule here:
$$B=\lim_{x\to0^+}\frac{(\ln x)'}{(\frac{1}{\sqrt x})'}$$
$$B=\lim_{x\to0^+}\frac{\frac{1}{x}}{\frac{(1)'\sqrt x-1\times(\sqrt x)'}{x}}$$
$$B=\lim_{x\to0^+}\frac{\frac{1}{x}}{\frac{0\times\sqrt x-\frac{1}{2\sqrt x}}{x}}$$
$$B=\lim_{x\to0^+}\frac{\frac{1}{x}}{\frac{-\frac{1}{2\sqrt x}}{x}}$$
$$B=\lim_{x\to0^+}\frac{1}{-\frac{1}{2\sqrt x}}$$
$$B=\lim_{x\to0^+}(-2\sqrt x)$$
$$B=-2\sqrt0$$
$$B=0.$$
Therefore, $$A=e^B=e^0=1$$
