Limit $\lim\limits_{x\to0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$ Without using L'Hopital's rule find:
$$\lim_{x\rightarrow 0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$$
I found the first derivative because I planned to use Taylor series:
\begin{align}
\frac{d}{dx}{\left(\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}\right)}
&=\frac{\left(\frac{1}{x+e}-e^x\right)\left(\cos^2{x} -e^x\right)-\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}
{(\cos^2{x} -e^x)^2}\\
&=\frac{\frac{1}{x+e}-e^x}{\cos^2{x}-e^x}-\frac{\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}{(\cos^2{x} -e^x)^2}
\end{align}
However, it seems I haven't gone so far. Should I start from the beginning and try a different method?
Source in Croatian: 2.kolokvij, matematička analiza

We have that
$$\frac{\ln{(e+x)}-e^x}x=\frac{\ln{(e+x)}-1}x-\frac{e^x-1}x \to\frac1e-1$$
$$\frac{\cos^2{x} -e^x}x=\frac{\cos^2 x-1}{x}-\frac{e^x-1}x=$$
$$=x(\cos x+1)\frac{\cos x-1}{x^2}-\frac{e^x-1}x \to 0\cdot 2\cdot \left(-\frac12\right)-1=-1$$
then
$$\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}=\frac{\ln{(e+x)}-e^x}{x}\cdot\frac{x}{\cos^2{x} -e^x}$$
