Can't solve this limit using $\frac{\sin x}x$ I've been told to solve the following limit using only $\lim_{x \to 0}\frac{\sin x}{x}=1$:
$$\lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}$$
I don't know how to do it except using L'Hospital's Rule but it's just insane amount of math and not the writer's intention(Symbolab do it but it's a computer so ;)
The answer is: $\frac18$
I'm looking for a way to solve this using $\frac{\sin x}x$.
 A: Appealing to the half-angle identity, $1-\cos(x)=2\sin^2(x/2)$, reveals 
$$\frac{1-\cos(1-\cos(x))}{x^4}=\frac{2\sin^2(\sin^2(x/2))}{x^4}$$
Next, we write
$$\begin{align}
\frac{2\sin^2(\sin^2(x/2))}{x^4}&=2\left(\frac{\sin(\sin^2(x/2))}{x^2}\right)^2\\\\
&=2\left(\frac{\sin(\sin^2(x/2))}{\sin^2(x/2)}\,\frac{\sin^2(x/2)}{x^2}\right)^2\\\\
&=2\,\color{blue}{\underbrace{\left(\frac{\sin^2(\sin^2(x/2))}{\sin^2(x/2)}\right)^2}_{\to 1}}\,\color{red}{\underbrace{\frac1{16}\left(\frac{\sin^2(x/2)}{(x/2)^2}\right)^2}_{\to 1/16}}\\\\
&\to \frac18
\end{align}$$
A: $$\begin{array}{rcl}
\displaystyle \lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}
&=&
\displaystyle \lim_{x \to 0} \frac{1-\cos^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{\sin^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{(1-\cos x)^2\sin^2(1-\cos x)}{x^4(1-\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos x)^2}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(\sin^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2 \left(\frac{\sin x}{x}\right)^4 \frac{1}{(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& 1^2 \cdot 1^4 \cdot \dfrac{1}{2^2 \cdot 2} \\
&=& \dfrac18
\end{array}$$
A: Don't forget $1-\cos x=2\sin^2(x/2)$. Therefore
$$1-\cos(1-\cos x)=2\sin^2\left(\sin^2\frac x2\right)
$$
and note
$$\frac{\sin(\sin^2 x/2)}{\sin^2 x/2}\to1$$
A: As $u\to 0,$
$$\frac{1-\cos u}{u^2} = \frac{1}{1+\cos u}\frac{1-\cos^2 u}{u^2} = \frac{1}{1+\cos u}\frac{\sin^2 u }{u^2}\to \frac{1}{2}\cdot 1^2 = \frac{1}{2}.$$
Therefore as $x\to 0,$
$$\frac{1-\cos (1-\cos x)}{x^4}= \frac{1-\cos (1-\cos x)}{(1-\cos x)^2}\frac{(1-\cos x)^2}{x^4}$$ $$ = \frac{1-\cos (1-\cos x)}{(1-\cos x)^2}\left (\frac{1-\cos x}{x^2}\right)^2 \to \frac{1}{2}\cdot \left ( \frac{1}{2}\right) ^2 = \frac{1}{8}.$$
