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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

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    Here's an idea: if you are allowed to use derivatives then you can establish polynomial bounds for the numerator and denominator (their Taylor polynomials to an $n$ and $n+1$st degree) and then use the squeeze theorem appropriately.2012-11-01

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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 $u={1\over6}$.

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    @alex.jordan: You are of course right.2012-11-01
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$\lim_{x\to 0}\frac{x^2-1+\cos^2x}{x^2}=\lim_{x\to 0}\frac{x^2-(1-\cos^2x)}{x^2}=\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2}=$ $=\lim_{x\to 0}(1-\frac{\sin^2x}{x^2})=1-\lim_{x\to 0}\frac{\sin^2x}{x^2}=1-\lim_{x\to 0}(\frac{\sin x}{x})^2=1-1=0$

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    Sorry but I made a considerable number of mistakes while writing the LATEX formula...2012-11-01
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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} $

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    @212121 Thanks! I have fixed the sign.2012-11-01
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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.

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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).

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).

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.