How to compute $\lim\limits_{x \to 0} \left(\frac{(\sin x)^2-x^2}{x^4}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
It's possible to resolve this limit with the developments of McLaurin?
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 0} \left(\frac{(\sin x)^2-x^2}{x^4}\right)$$
 A: You could very well use the series expansion of the function at $0$. Note that 
$$
\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\dots
$$
so 
$$
\sin^2(x)=x^2-\frac{x^4}{3}+\text{ higher order terms}
$$
so 
$$
\sin^2(x)-x^2=-\frac{x^4}{3}+\text{ higher order terms}
$$
so 
$$
\frac{\sin^2(x)-x^2}{x^4}=-\frac{1}{3}+\text{ higher order terms}
$$
so 
$$
\lim_{x\to0}\frac{\sin^2(x)-x^2}{x^4}=?
$$
A: $$\lim_{x\to 0}\frac{(\sin x)^2-x^2}{x^4}$$
$$=\lim_{x\to 0}\frac{\frac{1-\cos 2x}{2}-x^2}{x^4}$$
$$=\frac 12\lim_{x\to 0}\frac{1-\cos 2x-2x^2}{x^4}$$
Using L'Hospital's rule for $\frac{0}{0}$ form four times, one should get
$$=\frac 12\lim_{x\to 0}\frac{-2^4\cos 2x}{4\cdot 3\cdot 2\cdot 1}$$
$$=-\frac{16\cos 0}{48}=\color{red}{-\frac 13}$$
A: You can get there very easily by multiple application of the $\frac{0}{0}$ L'Hopital's rule:
$$\lim_{x \to 0} \frac{\sin^2 x - x^2}{x^4} = \lim_{x \to 0} \frac{\sin(2x) - 2x}{4x^3} = \lim_{x \to 0} \frac{2\cos(2x) - 2}{12x^2} = \lim_{x \to 0} \frac{-4\sin(2x)}{24x}  = \lim_{x \to 0} \frac{-8\cos(2x)}{24} = -\frac{1}{3}  $$
