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

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


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  • $\begingroup$ DId you at least write explicitly $\;\frac1{F(x)}\;$ ? Can you see some recognizable features there? And please follow the easy directions to write mathematics correctly in this site. $\endgroup$ – DonAntonio May 15 '16 at 20:34
  • 1
    $\begingroup$ I understand everything, my question is: what have you tried here? $\endgroup$ – DonAntonio May 15 '16 at 20:39
  • $\begingroup$ Can you use l'Hospital's rule? $\endgroup$ – DonAntonio May 15 '16 at 21:42

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


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    $\begingroup$ Nice, and fairly simpler than the direct application of L'Hôspital's rule. +1 $\endgroup$ – Jean-Claude Arbaut May 15 '16 at 23:18

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.


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 $-3$


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 $$\lim_{x\rightarrow0^+}-\left(\frac{x^2-\sin^2x}{x^2\sin^2x}\right)=\lim_{x\rightarrow0^+}-\frac{\left(\frac{x-\sin x}{x^3}\right)\left(\frac{x+\sin x}{x}\right)}{\left(\frac{\sin x}{x}\right)^2}=-\frac{\left(\frac16\right)(1+1)}{(1)^2}=-\frac13$$

  • $\begingroup$ Your first approach is correct, but the second approach has a limitation. It proves that if the limit in question exists then it has to be $-1/3$, but it does not prove that the limit exists. $\endgroup$ – Paramanand Singh May 16 '16 at 2:48
  • $\begingroup$ @Paramanand Singh: I only offered one approach. But I can see what your issue is. Maybe I can think of a way of fixing that potential problem without compromising the method too severely. $\endgroup$ – user5713492 May 16 '16 at 3:00
  • $\begingroup$ No it is not possible to fix this approach the limit is crucially dependent on the well known limit $\lim_{x \to 0}\dfrac{x - \sin x}{x^{3}}$ This can't be solved without L'Hospital or Taylor series. There is a very tedious approach which avoids L'Hospital and Taylor it uses infinite series. See this answer math.stackexchange.com/a/438121/72031 $\endgroup$ – Paramanand Singh May 16 '16 at 3:04
  • $\begingroup$ But +1 for the effort of proving that "if limit exists then it has to be $-1/3$." $\endgroup$ – Paramanand Singh May 16 '16 at 3:05
  • $\begingroup$ OK, @Paramanand Singh as promised I have rehabilitated the trisection method for your personal enjoyment. $\endgroup$ – user5713492 May 18 '16 at 9:15

$$\frac{1}{\sin^2 x}=\frac{1}{(x-x^3/6+O(x^5))^2}=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/6+O(x^4))^2}\\=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/3+O(x^4))}=\frac{1}{x^2}(1+x^2/3+O(x^4))$$


$$f(x)=\frac{1}{x^2}-\frac{1}{\sin^2 x}=-\frac13+O(x^2)$$


$$\frac{1}{f(x)}\underset{x\to 0}{\longrightarrow} -3$$

Without using Taylor expansion, you could apply L'Hôpital's rule to

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

However, you will need to differentiate four times, the numerator and the denominator (you get a $0/0$ form for lower derivatives). You will then get after simplification

$$\frac{\mathrm d^4 (x^2\sin^2x)}{\mathrm d x^4}=8(3-x^2)\cos(2x)-32x\sin(2x)$$ $$\frac{\mathrm d^4 (\sin^2x-x^2)}{\mathrm d x^4}=-8\cos(2x)$$



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