<s>[INST] 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 $\frac{\sin x}x$.
 [/INST] $$\begin{array}{rcl}
\displaystyle \lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}
&=&
\displaystyle \lim_{x \to 0} \frac{1-\cos^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{\sin^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{(1-\cos x)^2\sin^2(1-\cos x)}{x^4(1-\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos x)^2}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(\sin^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2 \left(\frac{\sin x}{x}\right)^4 \frac{1}{(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& 1^2 \cdot 1^4 \cdot \dfrac{1}{2^2 \cdot 2} \\
&=& \dfrac18
\end{array}$$
</s>