$\lim_{x\to 0}\frac{\sin^2x-x\tan x}{x^4}$

Question

fine the limit :

$$\lim_{x\to 0}\frac{\sin^2x-x\tan x}{x^4}$$

My Try :

$$\lim_{x\to 0}\frac{\sin^2x-x\tan x}{x^4}\\\lim_{x\to 0}\frac{\frac{\sin x \sin x}{x}-\frac{x\tan x}{x}}{x^4/x}\\\lim_{x\to 0}\frac{\sin x -\tan x}{x^3}=-1/2 $$

is it right ?

Answer 1

HINT:

$$\dfrac{\sin^2x-x\tan x}{x^4}=\dfrac{\sin x}{x\cos x}\cdot\dfrac{\sin x\cos x-x}{x^3}$$

Set $2x=y$ to get $$\lim_{x\to0}\dfrac{\sin x\cos x-x}{x^3}=\lim_{x\to0}\dfrac{\sin2x-2x}{2x^3}=4\cdot\lim_{x\to0}\dfrac{\sin y-y}{y^3}$$

Now see Are all limits solvable without L'Hôpital Rule or Series Expansion

Answer 2

Another hint, write \begin{equation} \frac{\sin^2 x - x\tan x}{x^4} = \frac{\sin^2 x}{x^2}\frac{\sin(2x) - 2x}{x^2\sin(2x)} \end{equation} In the limit $x \rightarrow 0$, $\sin(2x) - 2x = -4x^3/3$. You can confirm using Mathematica that your answer $-2/3$ is indeed correct.