Super power cos "sí"

$\begin{aligned} \lim_{x \to 0} (\cos x)^{\frac{1}{\sin^2 x}} & = \lim_{u \to 0} (\sqrt{1-u^2})^{\frac{1}{u^2}} \quad \quad \small \color{#3D99F6}{\text{Let } u = \sin x} \\ & = \lim_{u \to 0} (1-u^2)^{\frac{1}{2u^2}} \quad \quad \small \color{#3D99F6}{\text{Let } n = \frac{1}{2u^2} \quad \Rightarrow u^2 = \frac{1}{2n}} \\ & = \lim_{n \to \infty} \left(1 + \frac{\color{#3D99F6}{-\frac{1}{2}}}{n} \right)^n \\ & = e^{\color{#3D99F6}{-\frac{1}{2}}} \end{aligned}$

$\Rightarrow b = \boxed{-0.5}$

Let $f(x)=\cos(x)$ and $g(x)=\dfrac{1}{\sin^2(x)}$ (easier for me to type :P).

Since, $\displaystyle \lim_{x \to 0}{f(x)=1}$ and $\displaystyle \lim_{x \to 0}{(f(x)-1)g(x)}$ exists(it is equal to -1/2),

$\displaystyle \lim_{x \to 0}{f(x)^{g(x)}}=e^{\lim_{x \to 0}{(f(x)-1)g(x)}}=e^{-1/2}$ .

Therefore $b$ is $\boxed{{\dfrac{-1}{2}}}$ .

Let $y= \displaystyle \lim_{x \to 0} ( \cos x)^{\frac{1}{\sin^{2} x}}$ . Then $\ln y = \displaystyle \lim_{x \to 0} \dfrac{\ln \cos x}{\sin^{2} x}$

We have a limit in the form $\dfrac{0}{0}$ so we may apply L'Hopital's rule. That gives us $\ln y = \displaystyle \lim_{x \to 0} \dfrac{\dfrac{\sin x}{\cos x}}{2 \sin x \cos x}$

That reduces to $\displaystyle \ln y = \lim_{x \to 0 } \dfrac{-1}{2 \cos^{2} x}$ which becomes $\displaystyle \ln y = \dfrac{-1}{2}$ .

But what is $y$ ? Simply exponentiate both sides. $e^{\ln y} = y = \boxed{e^{\frac{ -1}{2}}}$

The exponent was required for the answer, which is simply $\boxed{-0.5}$ .