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)

$$\begin{eqnarray*} \frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)} 
& = & \underbrace{\frac{1}{e^x+\sqrt{1+2x+2x^2}}}_{\mbox{harmless}}\cdot \frac{e^{2x}-(1+2x+2x^2)}{x+\tan (x)-\sin (2x)} \\
& \stackrel{\mbox{Taylor}}{=} & \frac{1}{e^x+\sqrt{1+2x+2x^2}}\cdot \frac{\frac{4}{3}x^3+o(x^4)}{\frac{5}{3}x^3+o(x^4)} \\
&\stackrel{x \to 0}{\longrightarrow} & \frac{1}{2}\cdot \frac{4}{5} = \frac{2}{5}
\end{eqnarray*}$$
