Find the limit : $\lim _{x \to 0}x^{\sqrt{x}}=?$ Find the limit :
  $$\lim _{x \to 0}x^{\sqrt{x}}=?$$




My Try :

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

$$\ln f(x)= \sqrt{x}\ln x$$

Now what ?
You have added the tag limit-without-lhospital, therefore my answer is according to that.

Let $\sqrt{x}=u$. Thus your limit $$\lim_{x \to 0} \sqrt x \ln x= \lim_{u \to 0}   u \ln u^2= 2 \lim_{u \to 0} u \ln u $$

Thus, all you need to find is $\displaystyle \lim_{u \to 0^+}u\ln(u)$.

Let $u=e^{-t}$ and note that as $u \to 0^+$, we have $t \to \infty$. Hence,
$$L = \lim_{u \to 0} u \ln(u) = \lim_{t \to \infty} -te^{-t} = -\lim_{t \to \infty} \dfrac{t}{e^t}$$
Now recall that $e^t \geq \dfrac{t^2}2$. Hence, we have
$$\lim_{t \to \infty} \dfrac{t}{e^t} \leq \lim_{t \to \infty} \dfrac 2t = 0$$

Thus you get $$\color{red}{\lim_{x \to 0} \sqrt x \ln x= 0}$$
Use L'Hopital:

$$\lim_{x\to0} \sqrt{x}\ln x = \lim_{x\to0} \frac{\ln x}{\frac{1}{\sqrt{x}}} = \lim_{x\to0} \frac{\frac1x}{-\frac1{2x^{3/2}}} = -2\lim_{x\to 0} \sqrt{x} = 0$$

Now you have

$$\lim_{x\to 0} x^{\sqrt{x}} = \lim_{x\to 0} e^{\sqrt{x}\ln x} = e^0 = 1$$

using the continuity of exponentiation.