Is there a more efficient way of evaluating this limit?

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:

$\lim_{x\to 0} \frac{\sin(x)-\arctan(x)}{2x^3} = \lim_{x\to 0} \frac{\sin(x)-x + x -\arctan(x)}{2x^3}=\lim_{x\to 0} \frac{\sin(x)-x} {2x^3} + \lim_{x\to 0} \frac{x - \arctan(x)} {2x^3}$

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