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

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}

Robert Howard
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Archit
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    Would you show the steps you did? – Mark Sep 09 '18 at 09:31
  • 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 09 '18 at 09:34
  • Okay I'll try those. Thanks – Archit Sep 09 '18 at 09:43
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    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 09 '18 at 19:31
  • @Teepeemm Okay. Thanks that second part is what I really didn't know – Archit Sep 09 '18 at 19:33
  • You are not allowed to do the step from third to fourth line, once you have divided the expressions you need to apply l’Hopital separately for each term. – user Sep 09 '18 at 23:02
  • Evaluate the following comment about the title – Asaf Karagila Sep 10 '18 at 05:40
  • referring to the limit theorem, note that $\lim [f(x)\cdot g(x)]=\lim f(x) \cdot \lim g(x)$ if the limits on the right exist. When you separated the limit into two in line $3$, you got the first limit nonexisting ($\infty$), hence the separation is not successful. See Riemann's approach below. – farruhota Sep 10 '18 at 06:52
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    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
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    @JyrkiLahtonen I can't agree with you in that case. Indeed the limit itself is clearly a duplicate but the poser is not and indeed here we are dicussing in detail some issue related to the application of l'Hopital rule. Moreover I really can't see what is the problem with such kind of duplicates those are an enrichment for the community and since the duplicates are linked togheter the fact to have different answers from different user in different time and context can be very useful also for the future. Maybe you should consider carefully all these things before judging things negatively. Bye – user Sep 11 '18 at 22:25
  • See this answer. And this thread for more. – Jyrki Lahtonen Sep 12 '18 at 04:07
  • Sir. I get your point. But I wasn't wrong. And btw thanks for all the links they helped me too. I know I'll find alot of duplicates to this, I'm thankful for your insight, , now can we please let this go? – Archit Sep 12 '18 at 04:08
  • @ArchitJain Sorry I've reopened and then closed again the OP accidentally! – user Sep 12 '18 at 07:11
  • I reopened this post only as it is closed just due to an oversight by a user (see comment by @gimusi and the timeline). – quid Sep 12 '18 at 12:40
  • @quid Thanks and sorry for the inconvenience! – user Sep 12 '18 at 13:27

9 Answers9

11

By l'Hopital we have

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

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

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

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

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

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

user
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    Good steps! Had a hard job in typing. Great++ – Mikasa Sep 09 '18 at 09:53
  • @mrs I tried to apply l'Hopital pedantically to show that it is not a convenient method to solve limits without the knowledge of standard limits. Thanks – user Sep 09 '18 at 09:56
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    definitively you are masochist. +1 for typing this monstrosity :) – Masacroso Sep 09 '18 at 10:12
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    @Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P – user Sep 09 '18 at 10:14
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    I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1 – Paramanand Singh Sep 09 '18 at 11:45
  • @ParamanandSingh You know I was exactly thinking to you while typing that! Cheers – user Sep 09 '18 at 11:48
  • it would be easier (and L'Hospital is used thrice nicely) if it was transformed to: $\lim_\limits{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\lim_\limits{x \to 0}\frac{\sin^2 x-x^2}{x^4}$ – farruhota Sep 09 '18 at 12:00
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    @farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye – user Sep 09 '18 at 12:06
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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.

Mikasa
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10

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

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

user
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7

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

Riemann
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    Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that – Archit Sep 09 '18 at 09:45
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    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? – user Sep 09 '18 at 09:59
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    @gimusi and the final step also need some justification. – Masacroso Sep 09 '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. – user Sep 09 '18 at 10:15
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    just $\lim\limits_{x\to 0}\frac{\sin x+x}{x}=2$. – Riemann Sep 09 '18 at 10:22
  • +1 this is exactly what I would have written as answer. – Paramanand Singh Sep 09 '18 at 11:50
  • @Paramanand Singh: Thank you!! – Riemann Sep 09 '18 at 11:51
  • @Riemann What I mean is that the step should be written as $$\ldots =\lim_{x \to 0}\frac{(\sin x-x)(\sin x+x)}{x^4}=\lim_{x \to 0}\frac{(\sin x+x)}{x}\lim_{x \to 0}\frac{x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2x(\sin x-x)}{x^4}=\ldots$$ in order to do not give theimpression that we can always make these kind of substitution. – user Sep 09 '18 at 12:10
  • @ gimusi You are right. I edit it now! – Riemann Sep 09 '18 at 12:12
3

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:

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

Count Iblis
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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

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

user
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2

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):

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

1

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

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,

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

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

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

prog_SAHIL
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1

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 $\lim_{x \to 0}\frac{\sin x }x=1$.

user
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