Evaluate $\lim_\limits{x \to 0} (1 +\sin^2 x)^{\frac{1}{\ln(\cos x)}}$ $$\lim_{x \to 0} (1 + \sin^2 x)^{\frac{1}{\ln(\cos x)}}$$
I evaluated $\sin$ and $\cos x$ but what can be done with $\ln\left(1-\frac{x^2}{2}\right)$ or $\ln\left(\frac{2 - x^2}{2}\right)$?
Assume that L'Hopital is forbidden but you can use asymptotic simplifications like big and small $o$ notations and Taylor series.
 A: You can write the function as
$$(1 + \sin^2 x)^{ \frac{1}{\sin^2 x} \frac{\sin^2x}{\ln(\cos x)}}$$
Further
$$\frac{\sin^2x}{\ln(\cos x)}=\frac{x^2+o(x^2)}{\ln(1-\frac{x^2}{2}+o(x^2))}=\frac{x^2+o(x^2)}{-\frac{x^2}{2}+o(x^2)}\to-2$$
And 
$$(1 + \sin^2 x)^{ \frac{1}{\sin^2 x} } \to e$$
Hence...
A: L'Hôpital's rule is not needed for this limit. Here are the steps
$$\lim\limits_{x \to 0} \left(1 + \sin^2 x\right)^{\frac{1}{\ln(\cos x)}}$$
$$=\lim\limits_{x \to 0} \exp\left(\ln\left(1 + \sin^2 x\right)^{\frac{1}{\ln(\cos x)}}\right)$$
$$=\exp\left(\lim\limits_{x \to 0}\frac{\ln\left(1 + \sin^2 x\right)}{\ln(\cos x)}\right)$$
$$=\exp\left(2\lim\limits_{x \to 0}\frac{\ln\left(1 + \sin^2 x\right)}{\ln\left(1-\sin^2 x\right)}\right)$$
Let $h=\sin^2x$, then
$$\exp\left(2\lim\limits_{h\to 0}\frac{\ln\left(1 + h\right)}{\ln\left(1-h\right)}\right)$$
Now we can exploit the following equivalent infinitesimal
$$\ln(1+h)\sim h$$
Which leaves us with
$$\exp\left(2\lim\limits_{h\to 0}\frac{h}{-h}\right)=\exp\left(-2\right)$$
A: HINT: rewrite $$(1+\sin(x)^2)^{\frac{1}{\ln(\cos(x))}}$$ as $$e^{\frac{\ln(1+\sin(x)^2)}{\ln(\cos(x))}}$$
A: Let $y=\ln(\cos x)$ so $x=\cos^{-1}(e^y)$. The limit then becomes:
$$\lim_{y\rightarrow0}\left(2-e^{y}\right)^\frac{1}{y}$$
Applying a Taylor series you get:
$$\lim_{y\rightarrow0}\left(e^{-2}-4e^{-2}y+4e^{-2}y^3+...\right)$$
Which then gives the answer of: $e^{-2}$.
