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Math Help - Does l'hospitals rule apply here?

  1. #1
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    Does l'hospitals rule apply here?

    I'm asked to evaluate the limit as x approaches pi/2 from the left of sec(7x)cos(3x). Would l'hospitals rule apply here or am I going about this wrong?
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    Senior Member vincisonfire's Avatar
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    You can apply it anywhere but it is not necessary helpful.
     sec(7x)cos(3x) = \frac{cos(3x)}{cos(7x)}

     lim_{x=\frac{\pi}{2}} \frac{cos(3x)}{cos(7x)} = lim_{x=\frac{\pi}{2}} \frac{-3sin(3x)}{-7sin(7x)} = \frac{-3sin(3\frac{\pi}{2})}{-7sin(7\frac{\pi}{2})} = \frac{3}{7}
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    Well the reason I ask is that we just covered l'hospitals rule and this was on the homework we received, so I think perhaps we're supposed to use it if it applies
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    Senior Member vincisonfire's Avatar
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    Well it applies that's what I showed you.
    ( diff(cos(u)) = -sin(u)du)
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    Quote Originally Posted by vincisonfire View Post
    You can apply it anywhere but it is not necessary helpful.
     sec(7x)cos(3x) = \frac{cos(3x)}{cos(7x)}

     lim_{x=\frac{\pi}{2}} \frac{cos(3x)}{cos(7x)} = lim_{x=\frac{\pi}{2}} \frac{-3sin(3x)}{-7sin(7x)} = \frac{-3sin(3\frac{\pi}{2})}{-7sin(7\frac{\pi}{2})} = \frac{3}{7}


    And how did you do that last step, simplifying to just 3/7 when the two terms following 3 and 7 in the num. and denom. are different?
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  6. #6
    Newbie EightballLock's Avatar
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    Quote Originally Posted by fattydq View Post
    And how did you do that last step, simplifying to just 3/7 when the two terms following 3 and 7 in the num. and denom. are different?
    The value of \sin{\frac{a\pi}{2}}, assuming a is always a nonzero integer, will always equal 1, since sine is a periodic function. Go on, try it. So, he or she basically just canceled out 1 and 1 from the fraction.
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  7. #7
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    Ahh thanks to both of you
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  8. #8
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    Quote Originally Posted by EightballLock View Post
    The value of \sin{\frac{a\pi}{2}}, assuming a is always a nonzero integer, will always equal 1, since sine is a periodic function. Go on, try it. So, he or she basically just canceled out 1 and 1 from the fraction.
    Not quite... Depending on the remainder in the division of a by 4, \sin\frac{a\pi}{2} can take the values 0, 1 or -1. Go on, try it .
    In the present case, the explanation is that \frac{7\pi}{2}=\frac{3\pi}{2}+\frac{4\pi}{2}=\frac  {3\pi}{2}+2\pi and \sin(x+2\pi)=\sin x for any x (i.e. sine is periodic, with period 2\pi). (By the way, \sin\frac{3\pi}{2}=-1).
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