# 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

• Sorry, the coefficient of $sin$ is actually $x^3$ – 212121 Nov 1 '12 at 16:41
• Consider taking Taylor Expansions for both non-polynomial functions and see what happens. – busman Nov 1 '12 at 16:42
• What exactly is your question? Do you want to know how to reduce the original limit to $\lim_{x \to 0} \frac{x - \sin x}{x^3}$? Or, do you already know that, in which case your question is just how to determine the last limit is 1/6??? If it's just the second question, then this is an EXACT duplicate of math.stackexchange.com/q/217081/8671 – Graphth Nov 1 '12 at 17:58
• I guess this question isn't an EXACT duplicate because the previous one does not say we can assume the limit exists. So, this question is in fact easier, and the answer provided in the other question actually works for this one. On the other hand, the answer provided by Hans mentioned in that one works even if you don't assume the limit exists. – Graphth Nov 1 '12 at 18:08
• 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. – alex.jordan Nov 1 '12 at 18:26

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 $u={1\over6}$.

• +1 A fun argument; but it does assume that the limit exists in the first place. For example $u=\infty$ is another "solution" to the equation at the end. – alex.jordan Nov 1 '12 at 18:20
• @alex.jordan: You are of course right. – Christian Blatter Nov 1 '12 at 18:40

$$\lim_{x\to 0}\frac{x^2-1+\cos^2x}{x^2}=\lim_{x\to 0}\frac{x^2-(1-\cos^2x)}{x^2}=\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2}=$$ $$=\lim_{x\to 0}(1-\frac{\sin^2x}{x^2})=1-\lim_{x\to 0}\frac{\sin^2x}{x^2}=1-\lim_{x\to 0}(\frac{\sin x}{x})^2=1-1=0$$

• Look at the corrected question – 212121 Nov 1 '12 at 16:45
• while I work on answer question is changed – Adi Dani Nov 1 '12 at 16:49
• Sorry but I made a considerable number of mistakes while writing the LATEX formula... – 212121 Nov 1 '12 at 16:53

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.

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

• You made a mistake in the third equality. The sign is minus – 212121 Nov 1 '12 at 17:06
• The expression before $2/x^2$ should have had a difference in its numerator. – Benjamin Dickman Nov 1 '12 at 17:09
• @212121 Thanks! I have fixed the sign. – Sasha Nov 1 '12 at 18:12

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.