<s>[INST] Prove that $\lim\limits_{x\to 0}\frac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\frac{1}{6}$ without De L'Hôpital Rule or Taylor Expansion? Any ideas how to show $\lim\limits_{x\to 0}\dfrac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\lim\limits_{x\to 0}\dfrac{x-\sin(x)}{x^3}=\dfrac{1}{6}$ without using the De L'Hôpital rule (or proving a special case of it?).
How can we reduce this to $\lim\limits_{x\to 0}\dfrac{1-\cos(x)}{x^2}=\dfrac{1}{2}$?
You can suppose that we know the limit in question exists
and therefore use inequalities to bound it
 [/INST] $$\lim_{x\to 0}\frac{x^2-1+\cos^2x}{x^2}=\lim_{x\to 0}\frac{x^2-(1-\cos^2x)}{x^2}=\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2}=$$
$$=\lim_{x\to 0}(1-\frac{\sin^2x}{x^2})=1-\lim_{x\to 0}\frac{\sin^2x}{x^2}=1-\lim_{x\to 0}(\frac{\sin x}{x})^2=1-1=0$$
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