Is there a more efficient way of evaluating this limit? $$\lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{x^2\ln(1+2x+x^2)}$$
This was a multiple choice question so it should not take me such a long time but the only way I see how to attack it is just L'Hôpital rule multiple times which is so inefficient as it keeps getting bigger and I'm more likely to make errors. The limit is $ \frac{1}{12} $ but I'm not sure how to show it. If there are any helpful tips on how to reduce the amount of calculation and make it simpler, it'd be appreciated. 
 A: HINT
${\ln(1+2x+x^2)}=2\ln(1+x)$ and $\lim_{x\to 0} \frac{\ln (x+1)} x = 1$ therefore:
$\lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{x^2\ln(1+2x+x^2)} = \lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{2x^2\ln(1+x)} = \lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{2x^3 \frac {\ln(1+x)} x} = \lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{2x^3}  \lim_{x\to 0} \frac  1 {\frac {\ln (x+1)}{x}} = \lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{2x^3}$ 
then use L'Hospital or, even better:
$\lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{2x^3} = \lim_{x\to 0} \frac{\sin(x)-x + x -\arctan(x)}{2x^3}=\lim_{x\to 0} \frac{\sin(x)-x} {2x^3} + \lim_{x\to 0} \frac{x - \arctan(x)} {2x^3}$
A: Use Taylor's formula and equivalence of functions:


*

*$\sin x-\arctan x= x-\dfrac{x^3}6+o(x^3)-\Bigl(x-\dfrac{x^3}3+o(x^3)\Bigr)=\dfrac{x^3}6+o(x^3)\sim_0\dfrac{x^3}6$,

*$\ln(1+2x+x^2)=2\ln(1+x)=2x+o(x)\sim_02x$,
so that $$\frac{\sin(x)-\arctan(x)}{x^2\ln(1+2x+x^2)}\sim_0\frac{\dfrac{x^3}6}{x^2\cdot 2x}=\frac1{12}.$$

A: If you know about the Taylor development approach, you should be aware that the first order approximation of both $\sin x$ and $\arctan x$ are $x$. As both functions are odd, there is no quadratic term, and the difference will feature a cubic term if the cubic coefficients differ. So you can expect a numerator approximated by $ax^3$.
In the denominator, $\ln(1+y)$ yields $y$, so that $\ln(1+2x+x^2)$ yields $2x+x^2$ and other terms (starting from quadratic). Together with the factor $x^2$, the approximation is $2x^3$.
Hence all you need are the third order Taylor coefficients, which you might know by heart to be $-1/3!$ and $-1/3$, giving the limit
$$\frac{-\frac16+\frac13}2.$$

If you don't know the coefficients, derive three times
$$\sin x\to\cos x\to-\sin x\to-\cos x,$$ giving the Taylor coefficient $-1/3!$
For the arc tangent,
$$\arctan x\to\frac1{1+x^2}.$$ You can continue to evaluate the derivatives but it is easier to use the geometric series summation formula to write
$$\frac1{1+x^2}=1-x^2+\cdots$$ which integrates as
$$\arctan x=x-\frac{x^3}3+\cdots$$
A: You could replace
$\ln(1+2x+x^2)=2\ln(1+x)\sim2x$.
A: We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\sin x - \arctan x}{x^{2}\log(1 + 2x + x^{2})}\notag\\
&= \lim_{x \to 0}\frac{\sin x - \arctan x}{x^{2}\log(1 + x)^{2}}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\sin x - \arctan x}{x^{2}\log(1 + x)}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\dfrac{\sin x - \arctan x}{x^{3}\cdot\dfrac{\log(1 + x)}{x}}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\sin x - \arctan x}{x^{3}}\notag
\end{align}
Now you can either use L'Hospital's Rule or Taylor series approach (this one being easier) and easily get the answer as $1/12$.
Note that it is always a good idea to simplify the expression (whose limit is to be evaluated) via the use of identities (related to the functions involved) and standard limits like $\lim\limits_{x \to 0}\dfrac{\log(1 + x)}{x} = 1$ or $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ before applying the technique of L'Hospital's Rule or Taylor series.
