dessyj1
  • dessyj1
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Mathematics
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chestercat
  • chestercat
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dessyj1
  • dessyj1
\[\lim_{x \rightarrow \pi/2^-} \frac{ 2x-\pi }{ \cos^2(x) }\]
dessyj1
  • dessyj1
Is it -1 |dw:1450155733497:dw|
johnweldon1993
  • johnweldon1993
Hmm, what is the derivative of cos^2(x) again?

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dessyj1
  • dessyj1
oh dang
dessyj1
  • dessyj1
-2sin(x)cos(x)
johnweldon1993
  • johnweldon1993
Lets use the product rule to expand that out \[\large cos^2x = (cosx)(cosx)\] so product rule would give \[\large (-sinx)(cosx) + (cosx)(-sinx)\] or \[\large -2sinxcosx\] Which we know from trig identities is equal to \(\large -sin(2x)\) right?
dessyj1
  • dessyj1
alright
johnweldon1993
  • johnweldon1993
So after L'Hospital is applied we have \[\large \lim_{x \rightarrow \frac{\pi}{2}^-} -\frac{2}{sin(2x)}\] What would that get us from plugging in pi/2?
dessyj1
  • dessyj1
it would be negative inf in that case.
johnweldon1993
  • johnweldon1993
Correct indeed!
dessyj1
  • dessyj1
Thanks!
johnweldon1993
  • johnweldon1993
No problem!

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