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I need to find $$\lim_{x\to 0}\left(\frac{1}{1-\cos(x)}-\frac{2}{x^2}\right)$$ I already found it using Taylor series. However, I'm looking for a solution without Taylor series expansion or L'Hopital's rule because the problem was given in a calculus class at a point when only limits had been studied.

bjorn93
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  • It is a bit strange that this problem has been given at this stage since it seems not solveble without Taylor, lHopital or derivatives concept. – user Nov 17 '18 at 22:01
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    See the related OP – user Nov 17 '18 at 22:03
  • @KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor. – user Nov 17 '18 at 22:17
  • The limit involves evaluating $\lim_{x\to0}\frac{x-\sin x}{x^3}$ which could be done without Taylor or l'Hôpital, but in very convoluted ways. – egreg Nov 17 '18 at 22:19
  • @egreg Are you sure it can be done using $\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac16$? – user Nov 17 '18 at 22:22
  • @gimusi Yes, see Jack D'Aurizio's answer to the non duplicate question. – egreg Nov 17 '18 at 22:24
  • @egreg In the non duplicate question Jack D'Aurizio uses Taylor's. – user Nov 17 '18 at 22:31
  • @gimusi The key is when the limit is transformed into $\frac{1}{2}\lim_{z\to0}\frac{(z+\sin z)(z-\sin z)}{z^2\sin^2z}$. The denominator can be changed into $z^4$ and the fraction split as $\frac{z+\sin z}{z}\frac{z-\sin z}{z^3}$. The limit of the first fraction is elementary. – egreg Nov 17 '18 at 22:35
  • @egreg Ah ok, I was trying something similar but I couldn't obtain that. Put it as a solution, I think the OP is looking for that! For the convoluted limit refer here to L2 . – user Nov 17 '18 at 22:37
  • See https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Nov 18 '18 at 01:51

1 Answers1

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With the substitution $x=2z$, the limit becomes $$ \lim_{z\to0}\left(\frac{1}{\sin^2z}-\frac{1}{z^2}\right)= \frac{1}{2}\lim_{z\to0}\frac{z^2-\sin^2z}{z^2\sin^2z}= \frac{1}{2}\lim_{z\to0}\frac{z-\sin z}{z^3}\frac{z+\sin z}{z}\frac{z^2}{\sin^2z} $$ (see https://math.stackexchange.com/a/1357590/62967 for the idea about the substitution, not for the complete solution, that uses Taylor).

The second and third fractions have elementary limits $2$ and $1$ respectively. For the first fraction refer to https://math.stackexchange.com/a/1337564/62967

egreg
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  • The limits that are being used make use of Taylor expansions in their derivations. – herb steinberg Nov 17 '18 at 22:47
  • @herbsteinberg No it suffices to use $$\frac{z^2-\sin^2z}{z^2\sin^2z}=\frac{z^2-\sin^2z}{z^4}\frac{z^2}{\sin^2z}$$ and use that $\frac{z^2}{\sin^2z}\to 1$. – user Nov 17 '18 at 22:50
  • @herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+\frac{\sin z}{z}$, which doesn't need Taylor. – egreg Nov 17 '18 at 22:50
  • The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $\frac{sinz}{z}\to 1$ can be derived without Taylor. – herb steinberg Nov 17 '18 at 23:52
  • @herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $\sin z/z$ is one of them. – egreg Nov 18 '18 at 09:49
  • @egreg I see your solution. Do you think it can be done without using the limit of $\frac{z-\sin(z)}{z^3}$ (while still adhering to no L'Hopital or Taylor)? Perhaps more simple basic limits? I was under the impression this is a fairly straightforward problem. – bjorn93 Nov 18 '18 at 23:09
  • @bjorn93 I don't think so. Anyway, I consider these “no l'Hôpital” exercises just a waste of student's time, unless Taylor expansions are allowed. – egreg Nov 18 '18 at 23:12