Doubt in solving $\lim_{x\to 0}\cos(x)^{\frac{1}{\sin^2(x)}}$

To solve $$\lim_{x\to 0}\cos(x)^{\frac{1}{\sin^2(x)}}$$ I have done the following steps: $$\lim_{x\to 0}\cos(x)^{\frac{1}{\sin^2(x)}}=\lim_{x\to 0}(\sqrt{1-sin^2(x)})^{\frac{1}{\sin^2(x)}}=\lim_{x\to 0}(1-sin^2(x))^{\frac{-1}{-2\sin^2(x)}}=e^{-1/2}$$

My doubt is in the first equality: I mean, $\cos{x}=\pm\sqrt{1-\sin^2(x)}$, so why in this case can I choose the positive sign?

Solution 1:

$$\lim_{x\rightarrow 0}\left ( \cos x \right )^{1/\sin^2x}=\exp \left \{ \lim_{x\rightarrow 0}\frac{\ln \cos x}{\sin^2 x} \right \}=\exp \left \{ \lim_{x\rightarrow 0}\frac{\ln \cos x}{x^2} \right \}=\exp \left \{ \lim_{x\rightarrow 0}-\frac{x^2}{2}\cdot \frac{1}{x^2} \right \}=\frac{1}{\sqrt{e}}$$

1. I took the exponent and move out the degree per logarithm
2. I got a fraction, where I did not have addition and subtraction, therefore I can apply the equivalence
3. Notice, that $\sin \sim x$ and $\ln \cos x \sim -\frac{x^2}{2}$