L'Hôpital's rule exercise with $\sqrt{x}$ as exponent

Question

I'm a bit stuck trying to find the limit of the following function:

$$\lim_{x \to 0^+}\,\,{x^{\sqrt{x}}} $$

We are expected to use L'Hôpital's rule, and thus far I've managed to resolve the equation to the following form:

$$\lim_{x \to 0^+}\,\,{x^{\sqrt{x}}} = \lim_{x \to 0^+}\,\,{(e^{\ln{x}})^\sqrt{x}} = \lim_{x \to 0^+}\,\,{e^{\sqrt{x}\ln{x}}}$$

However, I can't get the equation to a fraction of the form $\frac{f(x)}{g(x)}$ to use the rule. Any suggestions on how I'd get started with this problem?

Thanks!

Edit: Using the advice provided, I've tried to evaluate the limit as follows:

$$\lim_{x \to 0^+}\,\,g(x) = \lim_{x \to 0^+}\,\,\ln{f(x)} = \lim_{x \to 0^+}\,\,\sqrt{x}\ln{x} = \lim_{x \to 0^+}\,\, \frac{\ln{x}}{x^{-\frac{1}{2}}}$$

Using L'Hôpital's rule:

$$\lim_{x \to 0^+}\,\, \frac{\ln{x}}{x^{-\frac{1}{2}}} = \lim_{x \to 0^+}\,\, \frac{\frac{1}{x}}{-\frac{1}{2}x^{-\frac{3}{2}}} = \lim_{x \to 0^+}\,\, \frac{1}{x}\cdot(-\frac{1}{2})\,x^{\frac{3}{2}} = \lim_{x \to 0^+}\,\, -2\sqrt{x}$$

According to this, the limit of $g(x)$ comes to $0$, so

$$\lim_{x \to 0^+}f(x) = e^0 = 1$$

Is this correct?

Answer 1

Hint $$\sqrt x\ln x=\frac{\ln(x)}{\frac{1}{\sqrt x}}$$

Answer 2

Write $$f(x)=x^{\sqrt x}$$

Then $$g(x)=\ln f(x)=\sqrt x\ln x=\frac{\ln x}{x^{-1/2}}$$

Now use l'Hopital to compute $$\lim_{x\to0^+}g(x)$$

Since $x\mapsto e^x$ is continuous,

$$\lim_{x\to 0^+}f(x)=e^{\lim_{x\to0^+}g(x)}$$