Proving: $\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac1{x^2}\right)=\frac{\pi ^2}3$ without L'Hospital

Question

Evaluating $$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$ with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to start.

Answer 1

Using Maclaurin expansion we get: $$ \frac{\sin\pi x}{\pi x} = \frac{\pi x - \frac16 (\pi x)^3 + O(x^5)}{\pi x} = 1 - \frac{\pi^2 x^2}{6} + O(x^4) \\ \frac{\pi x}{\sin\pi x} = 1 + \frac{\pi^2 x^2}{6} + O(x^4) \\ \left(\frac{\pi x}{\sin\pi x}\right)^2 = 1 + \frac{\pi^2 x^2}{3} + O(x^4) \\ \frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2} = \frac{1}{x^2}\left[\left(\frac{\pi x}{\sin\pi x}\right)^2-1\right] = \frac{\pi^2}{3} + O(x^2) \to \frac{\pi^2}{3} $$

Answer 2

Well, you've got your answer, and it's a good one, I'd use series expansions always in such case, but then, the answerer couldn't know you've ever heard of those expansions, and some of your comments show you aren't too familiar with them. That's why SE encourages sharing information about your mathematical background, btw. Most people ignore that. But then, you risk to get an answer like the following, without any l'Hospitals, based only on elementary principles:

"From the elementary identity $$\frac1{\sin^2x}-\frac1{x^2}=\sum^\infty_{k=1}3^{-2k}\,\frac{\frac83-\frac{16}9\sin^2\frac{x}{3^k}}{\left(1-\frac43\sin^2 \frac{x}{3^k}\right)^2},$$ letting $x\to0,$ we get $$\frac1{\sin^2x}-\frac1{x^2}\to\sum^\infty_{k=1}3^{-2k}\cdot\frac83=\frac13,$$ and the result we're looking for follows after replacing $x\to\pi x.$"

The joke is: that identity is an elementary consequence of the triplication formula $$\sin3y=3\sin y-4\sin^3y$$ and the limit $\sin y/y\to1$ as $y\to0,$ indeed.

Of course, such an answer isn't helpful, not just because it is rather obscure, but also because the method is applicable only in exceptional cases.

Answer 3

Pre-Calculus Answer to the Question

Note that since $\lim\limits_{x\to0}\frac{\sin(x)}x=1$, as shown in this answer, and $\frac1x$ is continuous at $x=1$, we also have $\lim\limits_{x\to0}\frac x{\sin(x)}=1$. $$ \begin{align} \lim_{x\to 0}\left(\frac{\pi^2}{\sin^2(\pi x)}-\frac1{x^2}\right) &=\lim_{x\to 0}\frac{\pi^2x^2-\sin^2(\pi x)}{x^2\sin^2(\pi x)}\tag{1a}\\ &=\lim_{x\to 0}\frac{\pi x-\sin(\pi x)}{(\pi x)^3}\lim_{x\to0}\frac{\pi x+\sin(\pi x)}{\sin(\pi x)}\lim_{x\to 0}\frac{\pi^2(\pi x)}{\sin(\pi x)}\tag{1b}\\[3pt] &=\frac16\cdot2\cdot\pi^2\tag{1c}\\[6pt] &=\frac{\pi^2}3\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: algebra
$\text{(1b)}$: factorization
$\text{(1c)}$: apply $\lim\limits_{x\to0}\frac x{\sin(x)}=1$ from above
$\phantom{\text{(1c):}}$ and $\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16$ from below
$\text{(1d)}$: computation

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Proof that $\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16}$

Assume that $0\lt x\le\frac\pi3$. Then, $\cos(x)\ge\frac12$ and $0\le\sin(x)\le x\le\tan(x)$. Therefore, $$ \begin{align} \frac{x-\sin(x)}{x^3} &\le\frac{\tan(x)-\sin(x)}{x^3}\tag{2a}\\ &=\frac{\tan(x)}{x}\frac{1-\cos(x)}{x^2}\tag{2b}\\ &=\frac1{\cos(x)}\frac{\sin(x)}{x}\frac{2\sin^2(x/2)}{4\,(x/2)^2}\tag{2c}\\[6pt] &\le1\tag{2d} \end{align} $$ Furthermore, $$ \begin{align} &\frac{x-\sin(x)}{x^3}-\frac14\frac{x/2-\sin(x/2)}{(x/2)^3}\tag{3a}\\ &=\frac{2(x/2)-2\sin(x/2)\cos(x/2)}{8(x/2)^3}-\frac{2(x/2)-2\sin(x/2)}{8(x/2)^3}\tag{3b}\\ &=\frac{2\sin(x/2)(1-\cos(x/2))}{8(x/2)^3}\tag{3c}\\ &=\frac{2\sin(x/2)\,2\sin^2(x/4)}{8(x/2)^3}\tag{3d} \end{align} $$ Since $\lim\limits_{x\to0}\frac{\sin(x)}x=1$, $(3)$ shows that $$ \lim_{x\to0}\left(\frac{x-\sin(x)}{x^3}-\frac14\frac{x/2-\sin(x/2)}{(x/2)^3}\right)=\frac18\tag4 $$ For any $n$, adding $\frac1{4^k}$ times $(4)$ with $x\mapsto x/2^k$ for $k$ from $0$ to $n-1$ gives $$ \begin{align} \lim_{x\to0}\left(\frac{x-\sin(x)}{x^3}-\frac1{4^n}\frac{x/2^n-\sin\left(x/2^n\right)}{\left(x/2^n\right)^3}\right) &=\frac18\frac{1-(1/4)^n}{1-1/4}\tag{5a}\\ &=\frac16-\frac16\frac1{4^n}\tag{5b} \end{align} $$ Thus, for any $\epsilon\gt0$, choose $n$ large enough so that $\frac1{4^n}\le\frac\epsilon2$. Then, $(5)$ says that we can choose a $\delta\gt0$ so that if $0\lt x\le\delta$, $$ \frac{x-\sin(x)}{x^3}-\overbrace{\frac1{4^n}\frac{x/2^n-\sin\left(x/2^n\right)}{\left(x/2^n\right)^3}}^{\frac12[0,\epsilon]_\#} =\frac16-\!\overbrace{\ \ \ \frac16\frac1{4^n}\ \ \ }^{\frac1{12}[0,\epsilon]_\#}\!+\frac12[-\epsilon,\epsilon]_\#\tag6 $$ where $[a,b]_\#$ represents a number between $a$ and $b$. The bounds above the braces follow from $(2)$ and the choice of $n$.

Equation $(6)$ says that for $0\lt x\le\delta$, $$ \frac{x-\sin(x)}{x^3}=\frac16+[-\epsilon,\epsilon]_\#\tag7 $$ Since $\frac{x-\sin(x)}{x^3}$ is even, we can say that $(7)$ is true for $0\lt|x|\le\delta$, which means that $$ \lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag8 $$

Answer 4

Yet another approach: as it's an even function, assume $x>0$. Cut a sector in a radius-$\sqrt{2}$ angle subtending $\pi x$ radians at the centre so $\pi x-\sin\pi x$ is the area in the sector outside the triangle with the same vertices. We'll approximate the arc as a parabola, in Cartesian coordinates with the line segment of the same endpoints a part of the $X$-axis, with extrema at $X=\pm\sqrt{2}\sin\frac{\pi x}{2}\sim\pm\frac{\pi x}{\sqrt{2}}$. The peak is at$$X=0,\,Y=\sqrt{2}(1-\cos\frac{\pi x}{2})=2\sqrt{2}\sin^2\frac{\pi x}{4}\sim\frac{\pi^2x^2\sqrt{2}}{8}.$$At leading order, the parabola is $Y=\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)$, so the area below it is$$\int_{-\pi x/\sqrt{2}}^{\pi x/\sqrt{2}}\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)dX=\frac{\pi^3x^3}{6}.$$So for small $x$,$$\pi x-\sin\pi x\sim\frac{\pi^3x^3}{6}\implies\frac{1}{\sin\pi x}-\frac{1}{\pi x}\sim\frac{\pi^2x^2}{6\sin\pi x}\sim\frac{\pi x}{6}\implies\frac{1}{\sin^2\pi x}-\frac{1}{\pi^2x^2}\sim\frac{\pi x}{2}\cdot\frac{2}{\pi x}=\frac13.$$