<s>[INST] Calculating $\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$ without L'Hospital's rule this is my first post here so excuse the lack of knowledge about how things usually go.
My question revolves around calculating the limit as $x$ approaches $0$ of the following function:
$$\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$$
The question came up in a test about a month ago and while I couldn't solve it in the test I've been working on it since then but I can't seem to get it. I know the limit is supposed to be $\frac{-2}{3}$ from some online calculators which abused l'hopital rule over and over again. I've tried playing around with it in so many ways but I always seem to get 0 over 0 or the so called indeterminate form. I've even tried calculating it by substituting in the Taylor series for the functions given but no luck. If anyone could show me a method of calculating this without using l'hopital rule or better yet, give me a hint as to how I should proceed I would be grateful.
 [/INST] More elementary solution (without Taylor series expansion or L'Hospital's rule) can be:
$$
\begin{align*}
\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{x^2}{1-\cos x}\cdot\frac{-(\cos x-1)}{\ln(\cos x)-0}\\
&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{x^2(1+\cos x)}{1-\cos^2x}\cdot\frac{-1}{\ln'(1)}\\
&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{2x^2}{\sin^2x}\cdot(-1)\\
&=-2\cdot\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}=L
\end{align*}
$$$$
\begin{align*}
L_1&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\\
&=\lim_{2x \to 0} \frac{\sin(4x^2)-\sin^2(2x)}{16x^4}=\lim_{x \to 0} \frac{2\sin(2x^2)\cos(2x^2)-4\sin^2(x)\cos^2(x)}{16x^4}\\
&=\lim_{x \to 0} \frac{4\sin(x^2)\cos(x^2)(1-2\sin^2(x^2))-4\sin^2(x)(1-\sin^2(x))}{16x^4}\\
&=\lim_{x \to 0} \frac{4\sin(x^2)(1-2\sin^2(x^2))-4\sin^2(x)(1-\sin^2(x))}{16x^4}\\
&=\lim_{x \to 0} \frac{4\sin(x^2)-4\sin^2(x)}{16x^4}-\lim_{x \to 0}\frac{8\sin^3(x^2)}{16x^4} +\lim_{x \to 0}\frac{4\sin^4(x)}{16x^4} \\
&=\frac{L_1}{4}-\lim_{x^2 \to 0}\frac{8}{16}\left(\frac{\sin x}{x}\right)^2\cdot\sin(x)+\lim_{x \to 0}\frac{4}{16}\left(\frac{\sin x}{x} \right)^4\\
&=\frac{L_1}{4}+\frac14\\
\therefore L_1&=\frac13
\end{align*}
$$
$$\therefore L=-2L_1=-\frac23$$
</s>