Finding the limit: $$\lim_{x \to 0}\frac {\sin x - \arctan x}{x^2 \ln x}$$
My questions:
1- I think the question should be corrected to as $x \rightarrow 0^+$, because of the domain of $\ln x$ ...... am I correct?
2- I applied L`hopital 2 times and after the second time it gaves me $0/-\infty $ which is $0$ .... am I correct?
Because of the domain of $\ln{x}$ the limit should approach as $x \rightarrow 0^{+} $
Solving the limit, on seeing the numerator the first thought is expansion but $\ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives
$$ \lim_{x \rightarrow 0^{+}}\frac{-\sin{x} + \frac{2x}{{1+x}^{2}}}{2\ln{x}+3} =0$$
Another way:
$$\dfrac{\sin x-\arctan x}{x^2\ln x}=\left(\underbrace{\dfrac{\sin x-x}{x^3}}-\underbrace{\dfrac{\arctan x-x}{x^3}}\right)\dfrac x{\ln x}$$
Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.
$$\lim_{x\to0^+}\dfrac x{\ln x}=\dfrac0{-\infty}=0$$
