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Vertical Asymptote

  1. Jan 29, 2006 #1
    Question: Find the Vertical Asymptote if any

    f(x)=[tex}\frac{2x}{\sin2x}[\tex]

    Im somewhat lost, I cross multi and get 1/2 although my friends tell me this just equals 1. Any ideas?

    err....cant get the cool math symbol things to work.....
     
    Last edited: Jan 29, 2006
  2. jcsd
  3. Jan 29, 2006 #2

    TD

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    Could you clarify the function? The LaTeX didn't work properly, check your tags.
     
  4. Jan 29, 2006 #3
    im trying to say f(x)= 2x / sin2x
     
  5. Jan 29, 2006 #4

    TD

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    Use the correct tags (closing with /, not \) and click to see the source:

    [tex]\frac{2x}{\sin2x}[/tex]

    Now, you can encounter a vertical asymptote if the denominator becomes zero for some value of x while the nominator is non-zero there.
    If both are zero, you can check with a limit what the function becomes.
     
  6. Jan 29, 2006 #5
    The only thing that would have the denom zero would be zero (sin2(o)) = 0
    but that would cause the nom to become zero
     
  7. Jan 29, 2006 #6

    TD

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    Both become zero there indeed, in which case you can't tell (yet) if there will be a vertical asymptote. Have you covered limits?
    You forgot one thing though, the sine-function becomes zero somewhere else (of course, it's period with period [itex]2\pi[/itex], but it has another zero within [0,[itex]2\pi[/itex]]). Think about the trigonometric unit circle, when is the sine zero as well?
     
  8. Jan 29, 2006 #7
    sin = zero @ pi (180* on unit circle)

    yep, covered limits. But in our notes he covers finding the VA without using limits (weird i know) so im not sure
     
  9. Jan 29, 2006 #8

    TD

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    Indeed, at [itex]x = \pi[/itex] as well! And the nominator doesn't become zero there, so we have a real vertical asymptote here.
    Of course, don't forget the sin(x) is periodic, so not only at [itex]x = \pi[/itex] but also at...? And the nominator only became zero at x = 0, but the denominator became zero there, so also at multiples of [itex]2 \pi[/itex] where the nominator doesn't become zero anymore!
    Can you follow?

    Perhaps you saw a standard limit you may use for the 0/0 case, which is the well-known:

    [tex]\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1[/tex]
     
  10. Jan 29, 2006 #9
    So my VA is x=pi ?
     
  11. Jan 29, 2006 #10

    TD

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    There's one yes, but there are more. Reread my last reply, I edited while you were posting.
     
  12. Jan 29, 2006 #11
    sin2x is zero when x=1/2 pi ? since that would mean sin = pi which would be zero on the denom
     
  13. Jan 29, 2006 #12

    TD

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    Well, sin(x) is a periodic function with period [itex]2\pi[/itex]. This means that if it becomes zero at x = 0, it also becomes zero when you add a multiple of [itex]2\pi[/itex]. The same goes for the case [itex]x = \pi[/itex].

    So we have that sin(x) becomes zero at: [itex]x = 0 + 2k\pi \vee x = \pi + 2k\pi [/itex] where k is an integer. Now, the nominator is non-zero in all those cases, except for x = 0 (but not for its multiples of [itex]2\pi[/itex]). So if we leave out x = 0 in the x values I listed, then we have all the places where a vertical asymptote occurs. You see?
     
  14. Jan 29, 2006 #13
    ok, i have you until the last sentsnce. if "k" is an interger, then the amount of VA's i could have is endless.

    Im also a little lost on what i should write down on my assignment

    by the way, thanks for all this help
     
  15. Jan 29, 2006 #14

    TD

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    Well, you're right! This function has an infinite number of vertical asymptotes!
     
  16. Jan 29, 2006 #15
    lol yea!

    now are you any good at continuity? I have a question

    using the def. of continuity to determine the points at which the function is no continous and indicate whether the discontinuity is removable or not

    f(x) = HUGE bracket (x+1)absolute value x-2 all over 2 sqrt x+1 x<0
    1 x=0
    -e^x / x+2 x>0

    I have no notes on continuity, so if you could just lead me to a good site if you dont have time to help then that fine
     
  17. Jan 29, 2006 #16
    and a quick question on limits

    lim x right arrow 0 e^3x -1 / e^x -1

    since anything to the root 0 is 1 you get 1-1/1-1 = 0
     
  18. Jan 29, 2006 #17

    TD

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    So you're looking for

    [tex]\mathop {\lim }\limits_{x \to 0} \frac{{e^{3x} - 1}}{{e^x - 1}}[/tex]

    Have you seen L'Hopital's rule? Or can you factor the nominator knowing that [itex]e^{3x} - 1 = \left( {e^x } \right)^3 - 1^3 [/itex]?
     
    Last edited: Jan 29, 2006
  19. Jan 29, 2006 #18
    L'Hopital's rule?
     
  20. Jan 29, 2006 #19

    TD

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    Oh I see you made a mistake: (1-1)/(1-1) is 0/0 which isn't the same as 0...! You encounter an 'indeterminate form' here, namely 0/0 where you can apply L'Hopital's rule which states that in this case, you can take the nominator's and denominator's derivative (seperately, so f/g becomes f'/g'). If you don't know this rule, try factoring so you'll be able to cancel out [itex]e^x-1[/itex].
     
  21. Jan 29, 2006 #20
    so like (e^x - 1)(e^x2 -e^x +1) so I now have e^x2 - e^x +1
     
    Last edited: Jan 29, 2006
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