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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \pi ^{+}}\frac{ \sin(\pi sinx)\sin \frac{ x }{ 4 } }{ \sqrt{1+cosx} } \]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0Not sure off the top of my head, maybe try L'Hospital's rule? You do have a 0/0 case...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0Or, maybe better, have you tried rationalizing the denominator? You might try to multiply top and bottom by \[\frac{\sqrt{1\cos x}}{\sqrt{1\cos x}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You might try the following identity on the numerator:\[\sin a \sin b = \frac{ 1 }{ 2 }\left[ \cos \left( ab \right)  \cos \left( a+b \right)\right]\]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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops. Looks like you still get zero. Sorry.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It might be helpful to substitute \(u = \pi  x \), in which case the limit becomes \[ \lim_{u\rightarrow 0} \frac{\sin(\pi \sin(u))\sin(\frac{\piu}{4})}{\sqrt{1\cos(u)}} \] 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 \[ \sin(\frac{\piu}{4}) \rightarrow \sin(\pi/4) \rightarrow 1/\sqrt{2} \]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0Not that computers can be trusted, but the numerical approximation appears to be heading towards pi as the limit...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Of course the substitution isn't necessary but it makes people feel better to take limits at zero. @jtvatsim yes, that's right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops... strictly speaking \[ \sqrt{u^2/2} = u/\sqrt{2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But there is no \( sin (\pi u) \) in the problem, is there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In the limit, you can replace the sin(u) with u. \[ \sin(\pi\sin(u)) \rightarrow \sin(\pi u) \rightarrow \pi u\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Righto. Forgot about that. Thanks.
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