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 $(x+1)$
 A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\lim_{x \to 0}{\expo{x}\cos\pars{x} - \pars{1 + x} \over \tan\pars{x} - \sin{x}}
=
\lim_{x \to 0}{x \over \tan\pars{x}}\,\bracks{x/2 \over \sin\pars{x/2}}^{2}{\expo{x} \cos\pars{x} - 1 - x \over x^{3}/2}
\\[3mm]&=
2\lim_{x \to 0}{\expo{x}\cos\pars{x} - x - 1 \over x^{3}}
=
2\lim_{x \to 0}{\expo{x}\cos\pars{x} - \expo{x}\sin\pars{x} - 1 \over 3x^{2}}
\\[3mm]&=
{2 \over 3}
\lim_{x \to 0}{\expo{x}\cos\pars{x} - \expo{x}\sin\pars{x} - \expo{x}\sin\pars{x} - \expo{x}\cos\pars{x} \over 2x}
=
-\,{2 \over 3}
\lim_{x \to 0}{\expo{x}\sin\pars{x} \over x}
\\[3mm]&=
-\,{2 \over 3}
\lim_{x \to 0}{\expo{x}\sin\pars{x} + \expo{x}\cos\pars{x} \over 1} =
\color{#0000ff}{\large -\,{2 \over 3}}
\end{align}
A: Just use L'Hopital 3 times, and you'll get the answer. The expressions only look long at first glance, but a bunch of stuff cancel out:

 $$\lim_{{x} \to {0}} \frac{e^x\cos x - x - 1}{tanx - sinx} L(\frac00)= \lim_{{x} \to {0}} \frac{e^x\cos x - e^x\sin x  - 1}{\sec^2 x - \cos x} L(\frac00)= \lim_{{x} \to {0}} \frac{-2e^x\ sinx}{\sin x + 2\tan x\sec^2 x} L(\frac00)= \lim_{{x} \to {0}} \frac{-2e^x(\ sinx + \cos x )}{\sec^4 x(4\sin^2 x + \cos^5 x + 2)} = -\frac23$$

A: The denominator goes to zero, numerator goes to 2, so the limit is easy to compute.
EDIT With the changed problem, the easiest way is to use power series. The denominator is asymptotic to $x^3/2,$ the numerator to $-x^3/3,$ so the limit should be $-3/2.$ But do the computation yourself.
A: $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{e^x\cos(x)-\left(x+1\right)}{\tan(x)\:-\sin(x)}\right)
& = \lim _{x\to 0}\left(\frac{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o\left(x^3\right)\right)\left(1-\frac{x^2}{2}+o\left(x^2\right)\right)-\left(x+1\right)}{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)\:-\left(x-\frac{x^3}{6}+o\left(x^3\right)\right)}\right)
\\& = \lim _{x\to 0}\left(\frac{\frac{-x^5-3x^4-4x^3}{12}+o\left(x^2\right)}{\frac{x^3}{2}+o\left(x^3\right)}\right)
\\& = \color{red}{-\frac{2}{3}}
\end{aligned}
$$
Solved with Taylor expansion
A: Note that we have $$\lim_{x\to 0}\frac{\tan x-\sin x} {x^{3}}=\lim_{x\to 0}\frac{\sin x} {x} \cdot \frac{1-\cos x} {x^{2}}\cdot\frac{1}{\cos x} =\frac{1}{2}$$ and therefore the desired limit is given by
\begin{align} 
L&=2\lim_{x\to 0}\frac{e^{x}\cos x - 1-x}{x^{3}}\notag\\
&=2\lim_{x\to 0}\left(\frac{\cos x - 1}{x^{2}}\cdot\frac{e^{x}-1}{x}+\frac{e^{x}+\cos x-x- 2 }{x^{3}}\right)\notag\\
&=2\left(-\frac{1}{2}+\lim_{x\to 0}\frac{e^{x}-\sin x-1}{3x^{2}}\right)\text{ (via L'Hospital's Rule)} \notag\\
&=-1+\frac{2}{3}\lim_{x\to 0}\frac{e^{x}-\cos x} {2x}\text{ (via L'Hospital's Rule)} \notag\\
&=-1+\frac{1}{3}\lim_{x\to 0}\left(\frac{e^{x}-1}{x}+x\cdot\frac{1-\cos x} {x^{2}}\right)\notag\\
&=-1+\frac{1}{3}=-\frac{2}{3}\notag
\end{align}
Note that before applying L'Hospital's Rule it is desirable to perform algebraic manipulation so that the process of differentiating the numerator and denominator is easy and leads to simpler expressions. 
A: l'hospital's rule is the right way to go. The first derivative will still give you $0/0$, but the second derivative should solve this problem.
