Evaluate the limit Squeeze Theorem Perhaps?

  • Thread starter crohozen
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  • #1
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Homework Statement


Evaluate the limit, if it exists:

lim[itex](\frac{sin x}{4x} * \frac{5-5 cos 3x}{2})[/itex]
x→0


The Attempt at a Solution



I understand that [itex]\frac{sin(x)}{4x}* \frac{4}{4} =\frac{1}{4}[/itex] but I don't know
what to do next because the 5-5cos3x/2 trips me up. I'm not seeing anything that I can do to it, so I'm thinking the Squeeze Theorem. Any help progressing further would be appreciated.
 

Answers and Replies

  • #2
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Start in pieces. Take the constants to the outside, and you're left with two terms.

[tex]\lim_{x\rightarrow 0} \left (\frac{\sin x}{4x}\cdot\frac{5-5\cos 3x}{2} \right) = \frac{5}{8}\lim_{x\rightarrow 0} \left ( \frac{\sin x}{x} - \frac{\sin x \cos 3x}{x} \right)[/tex]

See if you can take it from there.
 
  • #3
HallsofIvy
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I wouldn't divide it up quite the way process91 does. Look at
[tex]\frac{5}{8}\frac{sin x}{x}(1- cos(3x))[/tex]
and it should be obvious.

(If there had been an [itex]x^2[/itex] in the denominator rather than x, I would have made it
[tex]\frac{5}{8}\frac{sin(x)}{x}(3)\frac{1- cos(3x)}{3x}[/tex]
and it would be a bit more interesting!)
 
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  • #4
Ray Vickson
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You can use the fact that lim(f(x)*g(x)) = (lim f(x)) * (lim g(x)), if both limits on the right exist.

RGV
 

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