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

Use $1-\cos^2(x) = \sin^2(x)$:
$$
  \lim_{x\to 0} \frac{x^2-1+\cos^2(x)}{x^4 + x^3 \sin(x)} = \lim_{x\to 0} \frac{x^2 - \sin^2(x)}{x^4(1+\frac{\sin(x)}{x})} = \lim_{x\to 0} \frac{1 - \left(\frac{\sin(x)}{x}\right)^2}{x^2(1+\frac{\sin(x)}{x})} = \lim_{x\to 0} \frac{1 - \frac{\sin(x)}{x}}{x^2} = \frac{1}{6}
$$ 
