anonymous
  • anonymous
I have a question about limit.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\lim_{x \rightarrow \pi ^{+}}\frac{ \sin(\pi sinx)\sin \frac{ x }{ 4 } }{ \sqrt{1+cosx} } \]
abb0t
  • abb0t
@Empty
jtvatsim
  • jtvatsim
Not sure off the top of my head, maybe try L'Hospital's rule? You do have a 0/0 case...

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jtvatsim
  • jtvatsim
Or, maybe better, have you tried rationalizing the denominator? You might try to multiply top and bottom by \[\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}\]
anonymous
  • 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.
anonymous
  • anonymous
Oops. Looks like you still get zero. Sorry.
anonymous
  • 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 \[ \sin(\frac{\pi-u}{4}) \rightarrow \sin(\pi/4) \rightarrow 1/\sqrt{2} \]
jtvatsim
  • jtvatsim
Not that computers can be trusted, but the numerical approximation appears to be heading towards -pi as the limit...
anonymous
  • anonymous
Of course the substitution isn't necessary but it makes people feel better to take limits at zero. @jtvatsim yes, that's right.
anonymous
  • anonymous
Oops... strictly speaking \[ \sqrt{u^2/2} = |u|/\sqrt{2} \]
anonymous
  • anonymous
But there is no \( sin (\pi u) \) in the problem, is there?
anonymous
  • anonymous
In the limit, you can replace the sin(u) with u. \[ \sin(\pi\sin(u)) \rightarrow \sin(\pi u) \rightarrow \pi u\]
anonymous
  • anonymous
Right-o. Forgot about that. Thanks.

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