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
 A: A priori
$${x^2-1+\cos^2 x\over x^4+x^3\sin x}={x-\sin x\over x^3}\ .$$
Put $$\lim_{x\to0}{x-\sin x\over x^3}=:u\ .$$
Then from $\sin(3\alpha)=3\sin\alpha-4\sin^3\alpha$ we get
$${x-\sin x\over x^3}={1\over9}{{x\over3}-\sin{x\over3}\over\Bigl({x\over3}\Bigr)^3}+{4\over27}\left({\sin{x\over3}\over{x\over3}}\right)^3\ .$$
Here by definition of $u$ the right side converges to ${u\over9}+{4\over27}$ when $x\to0$. Therefore $u$ satisfies the equation $u={u\over9}+{4\over27}$ which has the unique solution $u={1\over6}$.
A: $$\lim_{x\to 0}\frac{x^2-1+\cos^2x}{x^2}=\lim_{x\to 0}\frac{x^2-(1-\cos^2x)}{x^2}=\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2}=$$
$$=\lim_{x\to 0}(1-\frac{\sin^2x}{x^2})=1-\lim_{x\to 0}\frac{\sin^2x}{x^2}=1-\lim_{x\to 0}(\frac{\sin x}{x})^2=1-1=0$$
A: Use $1-\cos^2(x) = \sin^2(x)$:
$$
  \lim_{x\to 0} \frac{x^2-1+\cos^2(x)}{x^4 + x^3 \sin(x)} = \lim_{x\to 0} \frac{x^2 - \sin^2(x)}{x^4(1+\frac{\sin(x)}{x})} = \lim_{x\to 0} \frac{1 - \left(\frac{\sin(x)}{x}\right)^2}{x^2(1+\frac{\sin(x)}{x})} = \lim_{x\to 0} \frac{1 - \frac{\sin(x)}{x}}{x^2} = \frac{1}{6}
$$ 
A: The way Sash noted above is correct. In fact:
$$
  \lim_{x\to 0} \frac{x^2-1+\cos^2(x)}{x^4 + x^3 \sin(x)} =  \lim_{x\to 0} \frac{1 - \frac{\sin(x)}{x}}{x^2}$$ but we should care that here $\sin(x)\approx x-x^3/6$ when $x$ is so close to zero. 
A: Squeeze Theorem.
First, there is some factoring and cancellation:
$$
\begin{align}
\frac{x^2-1+\cos^2x}{x^4+x^3\sin x} &=\frac{x^2-(1-\cos^2{x})}{x^3(x+\sin x)}\\
&=\frac{x^2-\sin^2(x)}{x^3(x+\sin x)}\\
&=\frac{(x-\sin x )(x+\sin x)}{x^3(x+\sin x)}\\
&=\frac{x-\sin x}{x^3}\\
\end{align}
$$
The function $f$ with $f(x)=\frac{1}{6}x^3$ and $g$ with $g(x)=x-\sin x$ satisfy the inequality $g(x)\leq f(x)$ for all $x\in(0,\epsilon)$. This is because at $x=0$, the two functions have the same value, derivative, second derivative, third derivative, and fourth derivative, but $g^{(5)}(0)<f^{(5)}(0)$. 
And if $h(x)=\frac{1}{6}x^3 - x^5$, then $h(x)\leq g(x)$ in $(0,\epsilon)$ for basically the same reason, but now $h^{(5)}(0)<g^{(5)}(0)$.
So for all $x\in (0,\epsilon)$, 
$$h(x)\leq g(x)\leq f(x)$$
$$\implies\frac{h(x)}{x^3}\leq \frac{g(x)}{x^3}\leq \frac{f(x)}{x^3}$$
$$\implies\frac{1}{6}-x^2\leq \frac{x-\sin x}{x^3}\leq \frac{1}{6}$$
Applying the Squeeze Theorem as $x\to0^+$ gives that $$\lim_{x\to0^+}\frac{x-\sin x}{x^3}=\frac{1}{6}$$ You can either alter this argument to work on both sides at once, or make a separate similar argument for the other side.
