# Evaluate $\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)$

$$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$

Using the L'Hospital Rule I obtained the value $$-1/4$$, but the answer is given to be $$-1/3$$. I can't find the mistake. Here's what I did; please point out the mistake.

\begin{align} \lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex] &=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex] &=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex] &=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex] &=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex] &=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex] &=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex] &=-\frac{1}{4} \end{align}

• Would you show the steps you did? – Mark Sep 9 '18 at 9:31
• Using L'Hôpital you need to apply at least twice. In this case is easier to use the asymptotic approximation $\sin x\sim_0 x$. As is said above: please, share your calculations. – Masacroso Sep 9 '18 at 9:34
• Okay I'll try those. Thanks – Archit Jain Sep 9 '18 at 9:43
• By the way, your mistake is that you can't split the limit into two pieces unless you know the separate limits exist. Having done that, you can't take $\sin x+x\cos x$ from one piece and put it into the other piece. – Teepeemm Sep 9 '18 at 19:31
• Approach0 finds dupes effortlessly. A downvote to every answer by a "trusted" user who didn't searh. – Jyrki Lahtonen Sep 11 '18 at 14:42

Hint: Write the function as $$\frac{\sin^2(x)-x^2}{x^4}\times \frac{x^2}{\sin^2(x)}$$ Otherwise use the Talor's expantion if you know it.

• And also it will help if you have memorized the "small angle approximation" $x/\sin(x) \rightarrow 1$ as $x \rightarrow 0$ – Mark Sep 9 '18 at 9:40
• I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it? – Archit Jain Sep 9 '18 at 9:43
• @ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler. – gimusi Sep 9 '18 at 9:57
• @mrs That's the good method to simplify by standard limits! – gimusi Sep 9 '18 at 9:58
• Yeah I got it now. Thanks – Archit Jain Sep 9 '18 at 9:58

By l'Hopital we have

$$\lim_{x \to 0}\frac{1}{x^2} - \frac{1} {\sin^2 x} =\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{\sin 2x-2x}{2x\sin^2 x+x^2\sin 2x }$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{2\cos 2x-2}{2\sin^2 x+2x\sin 2x+2x\sin 2x +2x^2\cos 2x}$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-4\sin 2x}{2\sin 2 x+8x\cos 2x+4 \sin 2x+4x\cos 2x-4x^2\sin 2x}$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-8\cos 2x}{12\cos 2 x+8\cos 2x-16x \sin 2x-8x\sin 2x+4\cos 2x-8x\sin 2x-8x^2\cos2x}$$

$$=\lim_{x \to 0}\frac{-8\cos 2x}{24\cos 2 x-32x \sin 2x-8x^2\cos2x} =\frac{-8}{24-0-0}=-\frac13$$

• Good steps! Had a hard job in typing. Great++ – mrs Sep 9 '18 at 9:53
• definitively you are masochist. +1 for typing this monstrosity :) – Masacroso Sep 9 '18 at 10:12
• @Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P – gimusi Sep 9 '18 at 10:14
• I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1 – Paramanand Singh Sep 9 '18 at 11:45
• @farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye – gimusi Sep 9 '18 at 12:06

As an alternative by Taylor expansion as $x\to 0$

$$\sin x = x -\frac16x^3 + o(x^3)\implies \sin^2 x = \left(x -\frac16x^3 + o(x^3)\right)^2=x^2-\frac13x^4+o(x^4)$$

we have

$$\frac{1}{x^2} - \frac{1} {\sin^2 x} =\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\frac{x^2-\frac13x^4+o(x^4)-x^2}{x^2\left(x^2-\frac13x^4+o(x^4)\right)}=$$$$=\frac{-\frac13x^4+o(x^4)}{x^4+o(x^4)}=\frac{-\frac13+o(1)}{1+o(1)}\to -\frac13$$

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

• Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that – Archit Jain Sep 9 '18 at 9:45
• The intermediate step $$\ldots =\lim_{x \to 0}\frac{(\sin x-x)(\sin x+x)}{x^4}$$ $$=\lim_{x \to 0}\frac{2x(\sin x-x)}{x^4}=\ldots$$ should be justified. Are you using taylor's expansion? – gimusi Sep 9 '18 at 9:59
• @gimusi and the final step also need some justification. – Masacroso Sep 9 '18 at 10:13
• @Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented. – gimusi Sep 9 '18 at 10:15
• just $\lim\limits_{x\to 0}\frac{\sin x+x}{x}=2$. – Riemann Sep 9 '18 at 10:22

My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $\sin(x)$. Using the Taylor expansion:

$$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} +\mathcal{O}(x^7)$$

Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $\mathcal{O}(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $\dfrac{1}{\sin^2(x)}$:

$$\frac{1}{\sin^2(x)} = \frac{1}{x^2}\left[1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^{-2}$$

To expand the square brackets, we can use:

$$\frac{1}{(1+u)^2} = 1-2 u + 3 u^2 + \mathcal{O}(u^3)$$

This can be derived by differentiating the geometric series term by term. We can then substitute $u = - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)$. We have:

$$u^2 = \left[- \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^2 = \frac{x^4}{36} +\mathcal{O}(x^6)$$

Therefore:

$$\frac{1}{1+u}= 1-2 u + 3 u^2 +\mathcal{O}(u^3)= 1 + \frac{x^2}{3} + \frac{x^4}{15} +\mathcal{O}(x^6)$$

And we see that:

$$\frac{1}{\sin^2(x)} = \frac{1}{x^2} + \frac{1}{3} + \frac{x^2}{15} +\mathcal{O}(x^4)$$

The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $\dfrac{1}{\sin^4(x)}$ by squaring both sides of this expansion, like:

$$\lim_{x\to 0}\left[\frac{1}{\sin^4(x)}-\frac{1}{x^4} - \frac{2}{3 x^2}\right]= \frac{11}{45}$$

• @gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative). – Count Iblis Sep 9 '18 at 17:07
• ops yes of course! – gimusi Sep 9 '18 at 17:22
• I've added also a solution by that way! – gimusi Sep 9 '18 at 19:29

As an alternative, following the idea by Count Iblis, we have that by Taylor expansion

$$\sin x = x-\frac16 x^3+o(x^3) \implies \frac1{\sin x}=\frac 1x\left(1-\frac16x^2+o(x^2)\right)^{-1}=\frac1x+\frac16x+o(x)$$

therefore

$$\left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) =\left( {\frac{1}{x}} + {\frac{1} {\sin x} }\right) \left( {\frac{1}{x}} - {\frac{1} {\sin x} }\right)=$$ $$=\left(\frac2x+\frac16x+o(x)\right) \left( -\frac16x+o(x)\right) =-\frac13+o(1) \to -\frac13$$

Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).

What you can do instead (notice the asymmetry):

$$\lim_{x\to0}\frac{\sin^2x-x^2}{x^2\sin^2x}=\lim_{x\to0}\frac{\sin^2x-x^2}{x^4}=\lim_{x\to0}\frac{\sin x+x}{x}\lim_{x\to0}\frac{\sin x-x}{x^3} \\=2\lim_{x\to0}\frac{\cos x-1}{3x^2}=-2\lim_{x\to0}\frac{\sin x}{6x}=-\frac13.$$

$$\lim_{x\to0}\frac{(\sin{x}+x)(\sin{x}-x)}{x\sin{x}\cdot x\sin{x}}$$

Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)

$$\lim_{x\to0}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}$$

$$\lim_{x\to0}\frac{x-\tan{x}}{x^3}=-\frac{1}{3}$$

$$\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$

So using this,

$$\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot \frac{(\sin{x}+x)}{x}\cdot \frac{(\sin{x}-x)}{x^3}$$

$$1\cdot2\cdot -\frac{1}{6}$$

$$-\frac{1}{3}$$

As noticed in the comments, we are allowed to proceed as follows

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

but we are not allowed to proceed as follows

$$\ldots=\lim_{x \to 0} \left( \frac{\sin x+x}{x\sin x} \right)\lim_{x \to 0}\left( \frac{\sin x-x}{x\sin x} \right)=\ldots$$

when one or both limits do not exist or the product leads to an undefined expression.

Notably in that case by l'Hopital we obtain

$$\ldots=\lim_{x \to 0} \frac {\cos x+1} {\sin x+x\cos x}\cdot \lim_{x \to 0} \frac {\cos x-1} {\sin x+x\cos x}=\ldots$$

and the LHS limit, in the form $\frac 2 0$, doesn't exist while the RHS limit is equal to zero.

Therefore the initial step in that case doesn't work.

Note that in any case also the following step

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

is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.

See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows

$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) = \lim_{x \to 0}\left(\frac{\sin^2 x-x^2}{x^4}\cdot\frac{x^2}{\sin^2 x}\right) \stackrel{?} = \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\ldots$$

and since, using l'Hopital for each part, we have

$$\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}=\lim_{x \to 0}\frac{\sin 2x-2x}{4x^3}=\lim_{x \to 0}\frac{2\cos 2x-2}{12x^2}=\lim_{x \to 0}\frac{-4\sin 2x}{24x}=\lim_{x \to 0}\frac{-8\cos 2x}{24}=-\frac13$$

$$\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\lim_{x \to 0}\frac{2x}{\sin 2x }=\lim_{x \to 0}\frac{2}{2\cos 2x }=1$$

we see that the initial step is allowed and then we can conclude that

$$\ldots= \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=-\frac13\cdot 1 =-\frac13$$

Note finally that some intermediate step can be highly simplified using the standard limit $\lim_{x \to 0}\frac{\sin x }x=1$.