calculate $\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$ 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?
 A: $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}\frac{\sqrt{1-\cos(\pi t)}}{\sqrt{1-\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)\sqrt{1-\cos(\pi t)}}{\sqrt{\sin^2(\pi t)}}$$
P.S. Pay attention to the sign of $\sin(\pi t)$ .
A: 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. 
A: 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.
A: 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) $
