<s>[INST] find the limit: $\lim_{x\to 0} \dfrac{e^x \cos x - (x+1)}{\tan x -\sin x}$ 
find
$\lim_{x\to 0} \dfrac{e^x \cos x - (x+1)}{\tan x -\sin x}$

i tried using l'hopital's rule, but it just gets very complicated very fast
edit: i made a mistake in the numerator (sorry!) its $(x+1)$
 [/INST] $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{e^x\cos(x)-\left(x+1\right)}{\tan(x)\:-\sin(x)}\right)
& = \lim _{x\to 0}\left(\frac{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o\left(x^3\right)\right)\left(1-\frac{x^2}{2}+o\left(x^2\right)\right)-\left(x+1\right)}{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)\:-\left(x-\frac{x^3}{6}+o\left(x^3\right)\right)}\right)
\\& = \lim _{x\to 0}\left(\frac{\frac{-x^5-3x^4-4x^3}{12}+o\left(x^2\right)}{\frac{x^3}{2}+o\left(x^3\right)}\right)
\\& = \color{red}{-\frac{2}{3}}
\end{aligned}
$$
Solved with Taylor expansion
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