$\lim\limits_{x\to 0}\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}$ 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)


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


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





The evaluating at $x=0$, you obtain


For the tangent, use

$$z=\tan x,$$






Finally, the limit is (by the three applications of L'Hospital)


  • $\begingroup$ In fact, the first computation is masochistically overkill, as the radical can be bypassed as shown by trancelocation. $\endgroup$ – Yves Daoust Nov 21 '18 at 16:19


By Taylor's series we have that

$$\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}=\frac{\overbrace{1+x+\frac12 x^2+\frac16x^3+o(x^3)}^{\color{red}{e^x}}-(\overbrace{1+x+\frac12x^2-\frac12 x^3+o(x^3)}^{\color{red}{\sqrt{1+2x+2x^2}}})}{x+\underbrace{x+\frac13 x^3o(x^3)}_{\color{red}{\tan x}}-(\underbrace{2x-\frac43 x^3 + o(x^3)}_{\color{red}{\sin 2x}})}$$

  • $\begingroup$ Rationalization then Taylor's expansion will be cleaner $\endgroup$ – lab bhattacharjee Nov 21 '18 at 14:56
  • $\begingroup$ @labbhattacharjee Yes that's also a nice hint to follow! Thanks $\endgroup$ – user Nov 21 '18 at 14:59

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



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