Calculate the following limit without L'Hopital $\lim\limits_{x\to 0}\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}$ I know how to count this limit with the help of l'Hopital rule. But it is very awful, because I need 3 times derivate it. So, there is very difficult calculations. I have the answer $\frac{2}{5}$. 
I want to know if there is other ways to calculate it, without 3 times using l'Hopital rule? (I could write my steps, but they are very big. I just took third derivative of numerator and denominator)
 A: $$\begin{eqnarray*} \frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)} 
& = & \underbrace{\frac{1}{e^x+\sqrt{1+2x+2x^2}}}_{\mbox{harmless}}\cdot \frac{e^{2x}-(1+2x+2x^2)}{x+\tan (x)-\sin (2x)} \\
& \stackrel{\mbox{Taylor}}{=} & \frac{1}{e^x+\sqrt{1+2x+2x^2}}\cdot \frac{\frac{4}{3}x^3+o(x^4)}{\frac{5}{3}x^3+o(x^4)} \\
&\stackrel{x \to 0}{\longrightarrow} & \frac{1}{2}\cdot \frac{4}{5} = \frac{2}{5}
\end{eqnarray*}$$
A: It is not so "very awful" as you say, if you tame the computation.
For the second term of the numerator, you can work this out as
$$y^2=1+2x+2x^2,$$
$$2yy'=2+4x,$$
$$2y'^2+2yy''=4,$$
$$4y'y''+2y'y''+2yy'''=0.$$
The evaluating at $x=0$, you obtain
$$y=1,y'=1,y''=1,y'''=-3.$$
For the tangent, use 
$$z=\tan x,$$
$$z'=z^2+1,$$
$$z''=2zz',$$
$$z'''=2z'^2+2zz''$$
giving
$$z=0,z'=1,z''=0,z'''=2.$$
Finally, the limit is (by the three applications of L'Hospital)
$$\frac{e^0-(-3)}{2-(-8)}.$$
A: HINT
By Taylor's series we have that
$$\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}=\frac{\overbrace{1+x+\frac12 x^2+\frac16x^3+o(x^3)}^{\color{red}{e^x}}-(\overbrace{1+x+\frac12x^2-\frac12 x^3+o(x^3)}^{\color{red}{\sqrt{1+2x+2x^2}}})}{x+\underbrace{x+\frac13 x^3o(x^3)}_{\color{red}{\tan x}}-(\underbrace{2x-\frac43 x^3 + o(x^3)}_{\color{red}{\sin 2x}})}$$
A: There is a simple way to address the denominator:
$$(x+\tan x-\sin2x)'=1+\tan^2x+1-2\cos 2x=\tan^2x+4\sin^2x=\sin^2x\left(\frac1{\cos^2x}+4\right).$$
The second factor will tend to $5$ and the first can be replaced by $x^2$.
Now using @trancelocation's trick and applying L'Hospital once, you need to find the limit of
$$\frac{2e^{2x}-(2+4x)}{2\cdot5x^2}.$$
