## anonymous one year ago I have a question about limit.

1. anonymous

$\lim_{x \rightarrow \pi ^{+}}\frac{ \sin(\pi sinx)\sin \frac{ x }{ 4 } }{ \sqrt{1+cosx} }$

2. abb0t

@Empty

3. jtvatsim

Not sure off the top of my head, maybe try L'Hospital's rule? You do have a 0/0 case...

4. jtvatsim

Or, maybe better, have you tried rationalizing the denominator? You might try to multiply top and bottom by $\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}$

5. anonymous

You might try the following identity on the numerator:$\sin a \sin b = \frac{ 1 }{ 2 }\left[ \cos \left( a-b \right) - \cos \left( a+b \right)\right]$That gives$\sin \left( \pi \sin x \right) \sin \frac{ \pi }{ 4 } = \frac{ 1 }{ 2 }\left[ \cos \left( \pi \sin x - \frac{ \pi }{ 4 } \right) - \cos \left( \pi \sin x + \frac{ \pi }{ 4 } \right)\right]$ This can be evaluated when $$x \rightarrow \pi$$. That leave the denominator.

6. anonymous

Oops. Looks like you still get zero. Sorry.

7. anonymous

It might be helpful to substitute $$u = \pi - x$$, in which case the limit becomes $\lim_{u\rightarrow 0-} \frac{\sin(\pi \sin(u))\sin(\frac{\pi-u}{4})}{\sqrt{1-\cos(u)}}$ Looking at the pieces separately as $$u\rightarrow 0$$, $\sin(u) \rightarrow u$ $\sin(\pi u) \rightarrow \pi u$ $\sqrt{1-\cos(u)} \rightarrow \sqrt{u^2/2} \rightarrow u/\sqrt{2}$ and $\sin(\frac{\pi-u}{4}) \rightarrow \sin(\pi/4) \rightarrow 1/\sqrt{2}$

8. jtvatsim

Not that computers can be trusted, but the numerical approximation appears to be heading towards -pi as the limit...

9. anonymous

Of course the substitution isn't necessary but it makes people feel better to take limits at zero. @jtvatsim yes, that's right.

10. anonymous

Oops... strictly speaking $\sqrt{u^2/2} = |u|/\sqrt{2}$

11. anonymous

But there is no $$sin (\pi u)$$ in the problem, is there?

12. anonymous

In the limit, you can replace the sin(u) with u. $\sin(\pi\sin(u)) \rightarrow \sin(\pi u) \rightarrow \pi u$

13. anonymous