Solving limit without L'Hôpitale rule or infinity serieses I've got this limit 
$$
{
\lim_{x\to 0} \left( \frac{e\cos(x)-e}{\ln(1+\ln(2-e^{sin^2(x)}))} \right)
\
}$$
I tried to remove NaN situation by 
$t=ln(2-e^{1-cos^2(x)}$).
But the problem was how can I know $1-cos(x)$ value .
So if there was any hint, please share it with me.
Thanks
 A: Rewrite the expression.
$${
 \frac{e\cos(x)-e}{\ln(1+\ln(2-e^{sin^2(x)}))}
}$$
$$=e\frac{\cos x-1}{x^2}\frac{\ln(2-e^{\sin^2 x})}{\ln(1+\ln(2-e^{\sin^2 x}))}
\frac{1-e^{\sin^2 x}}{\ln(2-e^{\sin^2 x})}\frac{\sin^2 x}{1-e^{\sin^2 x}}\frac{x^2}{\sin^2 x}$$
We see that the limit is $\frac{e}{2}$.
A: Let me first note that $ \lim\limits_{x \to 0} (\cos(x)+1) = 2$, or thus  $ \lim\limits_{x \to 0} \frac{\cos(x)+1}{2} = 1$.
Assuming the limit you posted exists and is finite, we have:
$$\begin{align}
\lim_{x\to 0}  \frac{e\cos(x)-e}{\ln(1+\ln(2-e^{\sin^2(x)}))} 
&= e \left(\lim_{x\to 0}  \frac{\cos(x)-1}{\ln(1+\ln(2-e^{\sin^2(x)}))} \right)\left(\lim_{x \to 0}\frac{\cos(x)+1}{2}\right)\\
&= \frac{e}{2} \lim_{x\to 0}  \frac{\cos^2(x)-1}{\ln(1+\ln(2-e^{\sin^2(x)}))} \\
&= -\frac{e}{2} \lim_{x\to 0}  \frac{\sin^2(x)}{\ln(1+\ln(2-e^{\sin^2(x)}))}\\
&= -\frac{e}{2} \lim_{t\to 0^+}  \frac{t}{\ln(1+\ln(2-e^t))} 
\end{align} \tag{1}
$$
where $t = \sin^2(x)$.
Now, the limit for $t$ you are left with can be solved in the way @ReneSchipperus showed in his answer:
$$\begin{align}\lim_{t\to 0^+}  \frac{t}{\ln(1+\ln(2-e^t))} &= \lim_{t\to 0^+}\left( \frac{\ln(2-e^t)}{\ln(1+\ln(2-e^t))} \cdot \frac{1-e^t}{\ln(2-e^t)} \cdot \frac{t}{1-e^t}\right)\\
&= \left(\lim_{t\to 0^+} \frac{\ln(2-e^t)}{\ln(1+\ln(2-e^t))}\right) \cdot \left( \lim_{t\to 0^+}\frac{1-e^t}{\ln(2-e^t)}\right) \cdot \left(\lim_{t\to 0^+} \frac{t}{1-e^t}\right)\\
&= \left(\lim_{u\to 0^+} \frac{u}{\ln(1+u)}\right) \cdot \left( \lim_{v\to 0^+}\frac{v}{\ln(1+v)}\right) \cdot \left(-\lim_{w\to 0^+} \frac{\ln(1+w)}{w}\right)\\
\end{align}\tag{2}$$
I substituted $u = \ln(2-e^t)$, $v = 1-e^t$ and $w = 1-e^t$.
Now, the limit 
$$\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1 = \lim_{x\to 0} \frac{x}{\ln(1+x)}$$
is well-known, hence the limit in $(2)$ becomes $-1$ and thus the limit in $(1)$ becomes $\frac{e}{2}$.
