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How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get

$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$ and this doesn't get me further. Any ideas?

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  • $\begingroup$ Tip: Multiply top and bottom with the denominator's conjugate and apply pythagorean theorem on the resulting denominator, as mentioned below....(I was too late) $\endgroup$ – imranfat May 21 '13 at 16:54
  • $\begingroup$ You could compute $\lim\limits_{t\rightarrow 1^-}{\sin^2(\pi t)\over {1+\cos(\pi t)}}$ instead. That's easy. Then note this limit is the negative of the square of your limit. $\endgroup$ – David Mitra May 21 '13 at 17:03
  • $\begingroup$ A bit of general advice concerning l'Hopital's Rule is that radicals are its "Achilles' heel". (The Rule should come with a warning on its label: Do Not Apply Directly to Radicals.) The problem is just what you found it does -- derivatives of radicals tend to remain such and simply volley back and forth between numerator and denominator. This is why we teach additional methods for dealing with indeterminate limit values. $\endgroup$ – colormegone May 21 '13 at 18:34
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$$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}\frac{\sqrt{1-\cos(\pi t)}}{\sqrt{1-\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)\sqrt{1-\cos(\pi t)}}{\sqrt{\sin^2(\pi t)}}$$

P.S. Pay attention to the sign of $\sin(\pi t)$ .

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  • $\begingroup$ well the sign is negative, but why is this important? $\endgroup$ – user31035 May 21 '13 at 17:00
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    $\begingroup$ @user31035 What is $\sqrt{\sin^2(\pi t)}$? $\endgroup$ – N. S. May 21 '13 at 17:01
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This is not quite a solution, more of a comment about the assertion that the L'Hospital's Rule calculation $$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$ does not get us any further.

Let $L$ be the original limit, assumed to exist and be non-zero. Now look at the right-hand side of the expression you reached. The $\pi$'s cancel. The term $\cos(\pi t)$ sedately approaches $-1$, cancelling the minus sign. And the rest has limit $\dfrac{1}{L}$! What you saw as a flaw becomes a virtue.

We conclude that $L=2\cdot \dfrac{1}{L}$. Thus $L=\pm \sqrt{2}$, and a quick examination of signs shows that we need the negative one, since a little past $\pi$, the sine is negative.

I do not advocate this approach, since there are details of existence to fill in, and one could easily reach an incorrect conclusion. Anyway, there is a simple non-L'Hospital calculation that quickly yields the answer.

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    $\begingroup$ I have seen more of those situations, where L'Hospital "loops you around", that an equation involving limit L can then be easily set up and solved. But yes, I would certainly make a graph to see if the answer makes sense. $\endgroup$ – imranfat May 21 '13 at 17:26
  • $\begingroup$ Often, when it looks as if we are going in circles, we are going in circles. But not always. $\endgroup$ – André Nicolas May 21 '13 at 20:19
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I'll show you a trick: since $t\to 1^+$ you know that $\pi t\to \pi^+$, so $(\pi t -\pi)\to 0^+$.

First of all do the change of variable $u=\pi t-\pi$, so you get $\pi t=u-\pi$. Then $\sin(\pi t)=\sin(u-\pi)=-\sin u$ and $\cos(\pi t)=\cos(u-\pi)=-\cos u$. Thus your limit becomes

$$ \lim_{u\to 0^+}\frac{-\sin u}{\sqrt{1-\cos u}}= -\lim_{u\to 0^+}\sqrt{\frac{\sin^2 u}{1-\cos u}}= -\lim_{u\to 0^+}\sqrt{1+\cos u}=-\sqrt{2} $$

The first equality is justified because $\sin u>0$ for $0<u<\pi$ (and you're interested in a right neighborhood of $0$).

Limits at zero are "psychologically" better, aren't they? There's really no difference with doing the limit at $\pi$ or using $\pi t$, but the region around $0$ is better known.

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In the original expression, use the half angle equation in the denominator, and the double angle equation in the numerator. Then $ cos(\pi t/2) $ cancels, and the original expression equals minus $\sqrt 2 sin(\pi t/2) $

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