How to calculate $\lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)^{-1}$? [closed]

$$f (x) = \frac{1}{x^2} - \frac{1}{\sin^2 x}$$

Find limit of $\dfrac1{f(x)}$ as $x\to0$.

Solution 1:

As already noted:

$$\frac1{f(x)}=\frac{x^2\sin^2x}{\sin^2x-x^2}=\color{blue}{\frac {x\sin^2x}{\sin x-x}}\cdot\color{purple}{\frac x{\sin x+x}}\;\color{red}{(**)}$$

Now, using l'Hospital's rule:

$$\begin{align*}&\lim_{x\to0}\color{blue}{\frac{x\sin^2x}{\sin x-x}}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\sin^2x+x\sin2x}{\cos x-1}\stackrel{\text{l'H}}=\lim_{x\to0}2\,\frac{\sin2x+x\cos2x}{-\sin x}\stackrel{\text{l'H}}=\\{}\\ &=\lim_{x\to0}2\frac{4\cos2x-2x\sin2x}{-\cos x}=2\frac{4}{-1}=-6\;,\;\;\;\;\text{whereas}\\{}\\ &\lim_{x\to0}\color{purple}{\frac x{\sin x+x}}=\lim_{x\to0}\frac1{\frac{\sin x}x+1}=\frac1{1+1}=\frac12\end{align*}$$

Thus, we finally get the limit is

$$\color{red}{(**)}=-6\cdot\frac12=-3$$

Solution 2:

You don't really need to work with $1/f(x)$; consider $$ f(x)= \frac{\sin^2x-x^2}{x^4}\frac{x^2}{\sin^2x}= \frac{\sin x-x}{x^3}\frac{\sin x+x}{x}\frac{x^2}{\sin^2x}= \frac{\sin x-x}{x^3}\left(\frac{\sin x}{x}+1\right) \left(\frac{x}{\sin x}\right)^2 $$ Since the limits of the second and third factors are well known, we just need to consider $$ \lim_{x\to0}\frac{\sin x-x}{x^3}= \lim_{x\to0}\frac{\cos x-1}{3x^2}= \lim_{x\to0}\frac{-\sin x}{6x}=-\frac{1}{6} $$ Therefore $$ \lim_{x\to 0}f(x)=-\frac{1}{6}\cdot 2\cdot 1^2=-\frac{1}{3} $$ and so $$ \lim_{x\to0}\frac{1}{f(x)}=\frac{1}{-1/3}=-3 $$

The limit above can be computed also with a Taylor expansion: $$ \lim_{x\to0}\frac{\sin x-x}{x^3}= \lim_{x\to0}\frac{x-x^3/6+o(x^3)-x}{x^3}=-\frac{1}{6} $$

Some knowledge of “basic” limits such as this one allows to use l'Hôpital more efficiently.

Solution 3:

Let's write explicitly

$${1\over f(x)}={x^2\sin^2{x}\over \sin^2{x}-x^2}$$

Then use the Taylor expansion $\sin{x}=x-x^3/6+o(x^4)$. We can write

$${1\over f(x)}={x^4+o(x^4)\over {-x^4\over 3}+o(x^4)}=-3+o(1)$$

And the limit we're looking for is $-3$

Solution 4:

This question is sort of a duplicate of a question that was asked earlier today in that $$\frac1{x^2}-\frac1{\sin^2x}=-\left(\frac{x^2-\sin^2x}{x^2\sin^2x}\right)$$ And there it was determined for $n\in\mathbb{Z}^+$, $$\lim_{x\rightarrow0}\frac{x^n-\sin^nx}{x^2\sin^nx}=\frac{n}{6}$$ Thus it follows from my analysis that $$\lim_{x\rightarrow0}\left(\frac1{x^2}-\frac1{\sin^2x}\right)=-\frac{2}{6}=-\frac1{3}$$ EDIT: The degree of complexity is so much less if $n=2$ that I thought I would write this case up separately. First off, from the angle sum and difference formulas $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$ we can derive $$\sin(n+1)x=\sin(nx+x)=\sin nx\cos x+\cos nx\sin x$$ $$\sin(n-1)x=\sin(nx-x)=\sin nx\cos x-\cos nx\sin x$$ Adding these two equations we find $$\sin(n+1)x+\sin(n-1)x=2\sin nx\cos x$$ And solving for $\sin(n+1)x$, we arrive at the key formula $$\sin(n+1)x=2\sin nx\cos x-\sin(n-1)$$ Since $\sin1x=\sin x$ and $\sin0x=\sin0=0$, $$\begin{align}\sin2x&=2\sin1x\cos x-\sin0x=2\sin x\cos x\\ \sin3x&=2\sin2x\cos x-\sin1x=2(2\sin x\cos x)\cos x-\sin x\\ &=4\sin x(1-\sin^2x)0\sin x=3\sin x-4\sin^3x\end{align}$$ And the fun may now begin! $$\begin{align}\frac1{x^2}-\frac1{\sin^2x}&=\frac{\sin^2x-x^2}{x^2\sin^2x}\\ &=\frac{(3(\sin(x/3)-(4/3)\sin^3(x/3)))^2-x^2}{x^2(3(\sin(x/3)-(4/3)\sin^3(x/3)))^2}\\ &=\frac{(\sin(x/3)-(4/3)\sin^3(x/3))^2-(x/3)^2}{9(x/3)^2(\sin(x/3)-(4/3)\sin^3(x/3))^2}\\ &=\frac{\sin^2(x/3)-(8/3)\sin^4(x/3)+(16/9)\sin^6(x/3)-(x/3)^2}{9(x/3)^2(\sin^2(x/3)-(8/3)\sin^4(x/3)+(16/9)\sin^6(x/3))}\\ &=\left(\frac19\frac{\sin^2(x/3)-(x/3)^2}{(x/3)^2\sin^2(x/3)}-\frac8{27}\frac{\sin^2(x/3)}{(x/3)^2}\right.\\&\left.+\frac{16}{81}\frac{\sin^4(x/3)}{(x/3)^2}\right)\frac1{1-(8/3)\sin^2(x/3)+(16/9)\sin^4(x/3)}\end{align}$$ Now we can take limits knowing that $$\lim_{x\rightarrow0}\sin(x/3)=0,\text{ and }\,\lim_{x\rightarrow0}\frac{\sin(x/3)}{(x/3)}=1$$ So $$\begin{align}\lim_{x\rightarrow0}\left(\frac1{x^2}-\frac1{\sin^2x}\right)&=\left(\frac19\lim_{x\rightarrow0}\left(\frac{\sin^2(x/3)-(x/3)^2}{(x/3)^2\sin^2(x/3)}\right)-\frac8{27}+0\right)\frac1{1-0+0}\\ &=\frac19\lim_{x\rightarrow0}\left(\frac1{x^2}-\frac1{\sin^2x}\right)-\frac8{27}\\ &=-\frac98\frac8{27}=-\frac13\end{align}$$ EDIT: @Paramanand Singh has pointed out that if one had no reason to believe that the limit existed in the first place, the above computation would not give one any more reason to believe it. I fixed this in my response but that proof was complicated and used angle bisection instead of trisection. Thus we start anew with (in what follows we assume that $0<x<\pi/4$) $$\begin{align}\sin x+\frac16\sin^3 x&=\left(3\sin(x/3)-4\sin^3(x/3)\right)+\frac16\left(3\sin(x/3)-4\sin^3(x/3)\right)^3\\ &=3\sin(x/3)+\frac12\sin^3(x/3)-18\sin^5(x/3)+24\sin^7(x/3)-\frac{32}3\sin^9(x/3)\\ &=3\left(\sin(x/3)+\frac16\sin^3(x/3)\right)-\frac23\sin^5(x/3)\left(16\cos^4(x/3)+4\cos^2(x/3)+7\right)\end{align}$$ We can rearrange this a little to $$\begin{align}3\left(\sin(x/3)+\frac16\sin^3(x/3)-(x/3)\right)-\left(\sin x+\frac16\sin^3 x-x\right)\\ =\frac23\sin^5(x/3)\left(16\cos^4(x/3)+4\cos^2(x/3)+7\right)<18\sin^5(x/3)\end{align}$$ Replacing $x$ by $x/3^{n-1}$ throughout and multiplying by $3^{n-1}$ we find that $$\begin{align}0&<3^n\left(\sin(x/3^n)+\frac16\sin^3(x/3^n)-(x/3^n)\right)\\ &-3^{n-1}\left(\sin(x/3^{n-1})+\frac16\sin^3(x/3^{n-1})-(x/3^{n-1})\right)\\ &<6\cdot3^n\sin^5(x/3^n)<6\frac{2^{5n}\sin^5x}{3^{4n}}=\frac{\frac{2^{5n}}{3^{4n}}6\sin^5x}{1-\frac{2^5}{3^4}}-\frac{\frac{2^{5n+5}}{3^{4n+4}}6\sin^5x}{1-\frac{2^5}{3^4}}\end{align}$$ That last inequality following because $$3\sin(x/3)=\sin x+4\sin^3(x/3)<2\sin x$$ For $0<x<\pi/4$ so that $$3^n\sin(x/3^n)<2^n\sin x$$ Now we can sum this inequality for $n=1$ to $n=N$ and it's a telescoping series so $$\begin{align}0&<3^N\left(\sin(x/3^N)+\frac16\sin^3(x/3^N)-(x/3^N)\right)-\left(\sin x+\frac16\sin^3x-x\right)\\ &<\frac{\frac{2^5}{3^{4}}6\sin^5x}{1-\frac{2^5}{3^4}}-\frac{\frac{2^{5N+5}}{3^{4N+4}}6\sin^5x}{1-\frac{2^5}{3^4}}<\frac{\frac{2^5}{3^{4}}6\sin^5x}{1-\frac{2^5}{3^4}}=\frac{192}{49}\sin^5x\end{align}$$ We can take the limit $$\begin{align}\lim_{n\rightarrow\infty}3^N\left(\sin(x/3^N)+\frac16\sin^3(x/3^N)-(x/3^N)\right)&=\lim_{n\rightarrow\infty}x\frac{\sin(x/3^N)+\frac16\sin^3(x/3^N)-(x/3^N)}{(x/3^N)}\\ &=x(1+0-1)=0\end{align}$$ So we know that $$0\le x-\sin x-\frac16\sin^3x\le\frac{192}{49}\sin^5x$$ Or $$0\le\frac{x-\sin x-\frac16\sin^3x}{x^3}\le\frac{192}{49}\frac{\sin^5x}{x^3}$$ Since $$\lim_{x\rightarrow0^+}\frac{192}{49}\frac{\sin^5x}{x^3}=\lim_{x\rightarrow0^+}\frac{192}{49}\left(\frac{\sin x}{x}\right)^3\left(\sin x\right)^2=\frac{192}{49}(1)^3(0)^2=0$$ It follows by the squeeze theorem that $$\lim_{x\rightarrow0^+}\frac{x-\sin x-\frac16\sin^3x}{x^3}=0$$ Then since $$\lim_{x\rightarrow0^+}\frac{\frac16\sin^3x}{x^3}=\lim_{x\rightarrow0^+}\frac16\left(\frac{\sin x}{x}\right)^3=\frac16(1)^3=\frac16$$ Then $$\lim_{x\rightarrow0^+}\frac{x-\sin x}{x^3}=\lim_{x\rightarrow0^+}\frac{x-\sin x-\frac16\sin^3x}{x^3}+\lim_{x\rightarrow0^+}\frac16\frac{\sin^3 x}{x^3}=0+\frac16=\frac16$$ So finally $$\lim_{x\rightarrow0^+}-\left(\frac{x^2-\sin^2x}{x^2\sin^2x}\right)=\lim_{x\rightarrow0^+}-\frac{\left(\frac{x-\sin x}{x^3}\right)\left(\frac{x+\sin x}{x}\right)}{\left(\frac{\sin x}{x}\right)^2}=-\frac{\left(\frac16\right)(1+1)}{(1)^2}=-\frac13$$