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
 A: 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
$$\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}=\frac{\ln{(e+x)}-e^x}{x}\cdot\frac{x}{\cos^2{x} -e^x}$$
A: Hint: show that $\frac {\ln (e+x)-e^{x}} x \to e^{-1}-1$ and $\frac {cos^{2}x -e^{x}} x \to -1$. Hence the limit is $(1-e^{-1})$.
A: 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.
A: I suggest using Taylor after some simplification:
\begin{eqnarray*}\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}
& = & \frac{1+\ln(1+\frac{x}{e}) - e^x}{\frac 12(1+\cos 2x)-e^x} \\
& = & \frac{1+\frac{x}{e}+o(x) - (1+x + o(x))}{\frac 12(1+ 1 +o(x))-(1+x+o(x))} \\
& = & \frac{\frac 1e -1 +o(1)}{-1+o(1)}\\
& \stackrel{x\to 0}{\longrightarrow}& 1-\frac 1e
\end{eqnarray*}
