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Finding a trigonometric limit

  1. Sep 20, 2009 #1
    This is from Thomas' Calculus Early Trancendentals, Media Upgrade, page 108, #27:

    The limit as x approaches 0 of (x csc(2x))/cos(5x)

    The answer is 1/2, which graphing confirms, but hell if I know how to get rid of the 5x and still come up with 1/2. It is boggling my mind. Any help would be appreciated.
     
  2. jcsd
  3. Sep 20, 2009 #2

    rock.freak667

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    csc(2x)=1/sin(2x)


    if I remember correctly

    [tex]\lim_{x \rightarrow 0} \frac{x}{sinx} =1[/tex]
     
  4. Sep 20, 2009 #3
    Yes, that much I know. The problem I have is with the cos(5x). How does one deal with that and come up with 1/2 as the final answer?
     
  5. Sep 20, 2009 #4
    Re-arrange then use L'Hopital

    [tex] \displaystyle\lim_{x\to 0} \frac{x\csc(2x)}{\cos(5x)} [/tex]

    [tex]= \displaystyle\lim_{x\to 0} \frac{x\csc(2x)}{1} [/tex]

    [tex]= \displaystyle\lim_{x\to 0} \frac{x}{\sin(2x)} \to \frac{0}{0}[/tex]

    [tex] = \displaystyle\lim_{x\to 0} \frac{1}{2\cos(2x)} = \frac{1}{2}[/tex]
     
    Last edited: Sep 20, 2009
  6. Sep 20, 2009 #5
    Well, we haven't covered L'Hopital yet. Also, the denominator is cos(5x), not 5x.
     
  7. Sep 20, 2009 #6
    You need to rewrite the limit to use other limits that you know, such as sinx/x

    [tex]\lim_{x \rightarrow 0} \frac{x\csc2x}{\cos5x} = \lim_{x \rightarrow 0} (x\csc2x * \frac{1}{\cos5x}) = \lim_{x \rightarrow 0} \frac{x}{\sin2x} * \lim_{x \rightarrow 0} \frac{1}{\cos5x}[/tex]

    [tex]= \lim_{x \rightarrow 0} \frac{1}{2} \frac{2x}{\sin2x} * \lim_{x \rightarrow 0} \frac{1}{\cos5x} = \frac{1}{2} \lim_{x \rightarrow 0} \frac{1}{\frac{\sin2x}{2x}} * \lim_{x \rightarrow 0} \frac{1}{\cos5x}[/tex]

    For [tex]\lim_{x \rightarrow 0} \frac{sin2x}{2x}[/tex]

    let k = 2x. Since 2x goes to 0 as k goes to 0,

    [tex]\lim_{x \rightarrow 0} \frac{\sin2x}{2x} = \lim_{k \rightarrow 0} \frac{\sin k}{k}[/tex]
     
    Last edited: Sep 20, 2009
  8. Sep 20, 2009 #7
    I made a typo in latex, if it were 5x then we'd be in trouble.

    L'Hopital

    [tex]\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)} = \displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}[/tex]

    if f(a)=g(a)=0 (or infinity)

    [tex]\frac{f(x)}{g(x)}=\frac{f(a)+(x-a)f'(a)+\frac{1}{2!}(x-a)^2g''(a)\cdots}{g(a)+(x-a)g'(a)+\frac{1}{2!}(x-a)^2g''(a)+\cdots}=\frac{(x-a)f'(a)+\frac{1}{2!}(x-a)^2g''(a)\cdots}{(x-a)g'(a)+\frac{1}{2!}(x-a)^2g''(a)+\cdots}=\frac{f'(a)+(x-a)g''(a)\cdots}{g'(a)+(x-a)g''(a)+\cdots}=\frac{f'(x)}{g'(x)}[/tex]

    f(a) and g(a) disappear and can divide top and bottom by 1/2(x-a).

    example

    [tex] \displaystyle\lim_{x\to 0} \frac{\sin (x)}{x} [/tex]

    Top and bottom tend to 0, so,

    [tex]\displaystyle\lim_{x\to 0} \frac{\sin (x)}{x} =\displaystyle\lim_{x\to 0} \frac{\cos (x)}{1} = 1 [/tex]
     
  9. Sep 20, 2009 #8
    I've read in my calc book and elsewhere that you shouldn't use L'Hopital's rule with sinx/x since that limit is used in proofs to find the derivative of sine. Circular reasoning...
     
  10. Sep 20, 2009 #9
    Thanks, Bohrok, that's exactly what I needed to set my brain straight.

    Perhaps if I would have remembered that the cosine of zero is one, I wouldn't have been in such a pickle. Hah.

    Thanks to everyone for their assistance.
     
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