Evaluate $\lim_\limits{x \to 0} (1 +\sin^2 x)^{\frac{1}{\ln(\cos x)}}$

Question

$$\lim_{x \to 0} (1 + \sin^2 x)^{\frac{1}{\ln(\cos x)}}$$

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.

Answer 1

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...

Answer 2

HINT: rewrite $$(1+\sin(x)^2)^{\frac{1}{\ln(\cos(x))}}$$ as $$e^{\frac{\ln(1+\sin(x)^2)}{\ln(\cos(x))}}$$

Answer 3

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: $e^{-2}$.

Answer 4

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 $$\exp\left(2\lim\limits_{h\to 0}\frac{h}{-h}\right)=\exp\left(-2\right)$$