Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0^+}x^{\sqrt x}=1$$
$$A=\lim_{x\to0^+}x^{\sqrt x}$$ 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, $$A=e^B=e^0=1$$