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

If you want the Taylor series, work separately numerator and denominator
$$\log (x+e)-e^x=\left(\frac{1}{e}-1\right) x-\frac{\left(1+e^2\right) }{2 e^2}x^2+\left(\frac{1}{3
   e^3}-\frac{1}{6}\right) x^3+O\left(x^4\right)$$
$$\cos ^2(x)-e^x=-x-\frac{3 }{2}x^2-\frac{1}{6}x^3+O\left(x^4\right)$$
$$\frac{\log (x+e)-e^x } { \cos ^2(x)-e^x}=\frac{\left(\frac{1}{e}-1\right) x-\frac{\left(1+e^2\right) }{2 e^2}x^2+\left(\frac{1}{3
   e^3}-\frac{1}{6}\right) x^3+O\left(x^4\right) } {-x-\frac{3 }{2}x^2-\frac{1}{6}x^3+O\left(x^4\right)}$$
Now, use the long division to get
$$\frac{\log (x+e)-e^x } { \cos ^2(x)-e^x}=\frac{e-1}{e}+\frac{1}{2} \left(-2+\frac{1}{e^2}+\frac{3}{e}\right) x+O\left(x^2\right)$$ which shows the limit and also how it is approached.
