Question

I have a question about limit.

Answer 1

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

Answer 2

@Empty

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

Oops. Looks like you still get zero. Sorry.

Answer 7

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
\[ \sin(\frac{\pi-u}{4}) \rightarrow \sin(\pi/4) \rightarrow 1/\sqrt{2} \]

Answer 8

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

Answer 9

Of course the substitution isn't necessary but it makes people feel better to take limits at zero.

@jtvatsim yes, that's right.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

Right-o. Forgot about that. Thanks.