Talk:One-sided limit

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I removed the following example:

  • We have
\lim_{x\downarrow 0}\arctan(x)={+\pi \over 2},
whereas
\lim_{x\uparrow 0}\arctan(x) = {-\pi \over 2}.

As far as I know, the arctangent is continuous everywhere on the real line, and this limit is zero. I would like to replace the example with one that does work, but I don't know any from trigonometry. It is true that lim x-> infty arctan x is pi/2, but one sided limits aren't very enlightening at infinity. If someone would like to explain what was supposed to be there, I'd like to have it back. -lethe talk 05:11, 18 November 2005 (UTC)

I've put it back with the needed correction. It now says the following:

  • We have
\lim_{x\downarrow 0}\arctan(1/x)={+\pi \over 2},
whereas
\lim_{x\uparrow 0}\arctan(1/x) = {-\pi \over 2}.

Michael Hardy 23:48, 18 November 2005 (UTC)

Now that you've said what it was supposed to be, it should have been obvious. Anyway, thanks. -lethe talk 00:10, 19 November 2005 (UTC)
I found it a bit strange on Michael's behalf to remove an elementary example with a picture, in favor of an example which duplicates an existing one, that is containing 1/x with x->0.
By no means do I plan it as a revenge, but after Lethe put my example back, I removed Michael's second example. Now we have two very different and somewhat representative examples, as things should be. Oleg Alexandrov (talk) 00:38, 19 November 2005 (UTC)