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.

  • $\begingroup$ 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. $\endgroup$ – gimusi Nov 17 '18 at 22:01
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    $\begingroup$ See the related OP $\endgroup$ – gimusi Nov 17 '18 at 22:03
  • $\begingroup$ @KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor. $\endgroup$ – gimusi Nov 17 '18 at 22:17
  • $\begingroup$ 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. $\endgroup$ – egreg Nov 17 '18 at 22:19
  • $\begingroup$ @egreg Are you sure it can be done using $\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac16$? $\endgroup$ – gimusi Nov 17 '18 at 22:22

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

  • $\begingroup$ The limits that are being used make use of Taylor expansions in their derivations. $\endgroup$ – herb steinberg Nov 17 '18 at 22:47
  • $\begingroup$ @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 $\frac{z^2}{\sin^2z}\to 1$. $\endgroup$ – gimusi Nov 17 '18 at 22:50
  • $\begingroup$ @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. $\endgroup$ – egreg Nov 17 '18 at 22:50
  • $\begingroup$ 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. $\endgroup$ – herb steinberg Nov 17 '18 at 23:52
  • $\begingroup$ @herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $\sin z/z$ is one of them. $\endgroup$ – egreg Nov 18 '18 at 9:49

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