How can I calculate $\lim\limits_{x\to 0} \frac {(\sin(x)-\tan(x))^2}{(e^x-1-\ln(x+1))^3}$? 
How can I calculate $\lim\limits_{x\to 0} \dfrac {(\sin(x)-\tan(x))^2}{(e^x-1-\ln(x+1))^3}$?

I have tried really hard to solve this limit, but I cannot solve it. The answer is $1/4$, but I want to learn how to do it, because learning is my goal. Thanks for your time
 A: The simplest way to handle a limit of this form is using Taylor series.
We know that in a neighborhood of $x=0$:


*

*$\sin x=x-\frac16x^3+O(x^5)$

*$\tan x=x+\frac13x^3+O(x^5)$

*$e^x=1+x+\frac12x^2+O(x^3)$

*$\ln(x+1)=x-\frac12x^2+O(x^3)$


So $\sin x-\tan x=-\frac12x^3+O(x^4)$ and $(\sin x-\tan x)^2=\frac1{4}x^6+O(x^7)$.  Likewise, $e^x-1-\ln(x+1)=x^2+O(x^3)$ and $(e^x-1-\ln(x+1))^3=x^6+O(x^7)$, so the limit in question is $\lim\limits_{x\to0}\dfrac{\frac14x^6+O(x^7)}{x^6+O(x^7)}=\lim\limits_{x\to 0}\left(\frac14+O(x)\right)=\frac14$.
A: Take the parts separately, evaluate 
$$\lim\limits_{x\to 0}\frac{\sin x-\tan x}{x^3}$$
this can be rewritten as
$$\frac{\sin x-\tan x}{x^3}=\frac{\sin x}{x}\frac{1}{\cos x}\frac{\cos x-1}{x^2}\rightarrow -\frac{1}{2}$$
Then find
$$\lim\limits_{x\to 0}\frac{e^x-1-\ln(x+1)}{x^2}$$
this can also be expressed in terms of standard limits
$$\frac{e^x-1-\ln(x+1)}{x^2}=\frac{e^x-1-x}{x^2}+\frac{x-\ln(x+1)}{x^2}\rightarrow \frac{1}{2}+\frac{1}{2}=1$$
Put it all together you get $\frac{1}{4}$.
A: Here is a way to do it without Taylor series.
Start by noting that
$$(\sin x-\tan x)^2={\sin^2x(1-\cos x)^2\over\cos^2x}={\sin^2x(1-\cos x)^2(1+\cos x)^2\over\cos^2(1+\cos x)^2}={\sin^6x\over\cos^2x(1+\cos x)^2}$$
and that $\cos^2x(1+\cos x)^2\to4$ as $x\to0$, so that
$$\lim_{x\to0}{(\sin x-\tan x)^2\over(e^x-1-\ln(x+1))^3}={1\over4}\lim_{x\to0}\left(\sin^2x\over e^x-1-\ln(x+1) \right)^3$$
Thus it suffices to show that
$$\lim_{x\to0}{\sin^2x\over e^x-1-\ln(x+1)}=\lim_{x\to0}\left(\sin x\over x \right)^2\lim_{x\to0}{x^2\over e^x-1-\ln(x+1)}=1$$
The limit for $\sin x\over x$ should be familiar.  The other limit is easily done with two rounds of L'Hopital:
$$\lim_{x\to0}{x^2\over e^x-1-\ln(x+1)}=\lim_{x\to0}{2x\over e^x-{1\over x+1}}=\lim_{x\to0}{2\over e^x+{1\over(x+1)^2}}={2\over1+1}=1$$
