Solving limit $\lim_{x\to0^+}\cos(\sqrt{x})^{1/x}$ without l'Hospital's rule

Question

How to solve this limit $$\lim_{x\to0^+}\cos(\sqrt{x})^{1/x}$$ without L'Hospital's rule.

Answer 1

HINT

Use that

$$\cos(\sqrt{x})^{1/x}=\left[(1+(\cos(\sqrt{x})-1))^{\frac{1}{\cos(\sqrt{x})-1}}\right]^\frac{\cos(\sqrt{x})-1}{x}$$

and the relevant standard limits.

Answer 2

Using the series expansion of $~\cos~$ (Taylor series at $x=0$):

$\displaystyle\cos(\sqrt{x})^{1/x}\approx\left(1-\frac{x}{2}\right)^{1/x}\approx e^{-1/2}$

Answer 3

Let $\sqrt x=2y$

so that the limit becomes $$\left(\lim_{y\to0^+}(1-2\sin^2y)^{-1/2\sin^2y}\right)^{-2\sin^2y/4y^2}$$

$$=e^{-1/2}$$