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A limit problem

  1. Jul 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Lim (4t2)*(sin(2/t))2
    t→∞


    2. Relevant equations



    3. The attempt at a solution
    I know it will be a indeterminant form (∞*0) if I don't do anything to the expression, but I don't how.
     
  2. jcsd
  3. Jul 9, 2013 #2

    vanhees71

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    2016 Award

    Taylor expansion of the sine helps!
     
  4. Jul 9, 2013 #3
    I am in Cal I:frown:
     
  5. Jul 9, 2013 #4
    Hint: [itex] lim_{t \rightarrow \infty} t \sin(1/t)= lim_{x \rightarrow 0} \frac{sin(x)}{x}=1 [/itex] by making the substitution x=1/t. Use this combined with the double angle formula for sine to compute your limit.
     
  6. Jul 9, 2013 #5
    Why did you change t→∞ to x→0? I don't understand this.
     
  7. Jul 9, 2013 #6
    If x=1/t then x goes to 0 as t becomes large.
     
  8. Jul 9, 2013 #7

    micromass

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    Have you seen L'Hospitals rule?
     
  9. Jul 9, 2013 #8
    I still can't do it :(
     
  10. Jul 9, 2013 #9

    Dick

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    Probably because you didn't pay much attention to HS-Scientist's fine suggestion. Try it. Change the variable to x where x=1/t. So if t->infinity then x->0.
     
  11. Jul 9, 2013 #10
    Okay. Here is how I tried.
    Lim (4*(sin x)2)/x2 because no much difference between 2/t and 1/t
    X→0

    Then I got an indeterminant form 0/0, so I used L'hopital's rule.
    Lim (8sin x cos x)/(2x)
    X→0

    Still 0/0 ,so again
    Lim (8(cosx)2-8(sinx)2)/2
    X→0
    Which equals 4.
    But my calculator says the answer should be 16:confused:
     
    Last edited: Jul 9, 2013
  12. Jul 10, 2013 #11

    Mark44

    Staff: Mentor

    Both 2/t and 1/t approach 0 as t gets large, but you can't ignore the difference.

    In your original limit, you had sin2(2/t), so what does that become with the substitution x = 1/t?
     
  13. Jul 10, 2013 #12
    Right! That is what made me wrong. Thank you so much!
    But I got the answer without using double-angle formula. Just keep applying L'hopital's rule:)
     
  14. Jul 10, 2013 #13

    Mark44

    Staff: Mentor

    You don't need either the double-angle formula or L'Hopital's Rule if you use this limit:
    $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
    This was suggested by HS-Scientist back in post #4.
     
  15. Jul 10, 2013 #14
    $$\lim_{x \to 0} \frac{4\sin^2(2x)}{x^2}$$

    How do you use that limit to get 16?
    This question comes from the AP test. So if I were to use the way I did to solve it , it would waste lots of time.
     
  16. Jul 10, 2013 #15
    $$\lim_{x \to 0} \frac{4\sin^2(2x)}{x^2}=4 (\lim_{x \to 0} \frac{sin(2x)}{x})^2=4(2^2)=16 $$

    Believe me, you would have enough time on the AP test to do this. I took it just this year.
     
  17. Jul 10, 2013 #16
    I didn't know you could just factor 2 out of the sin. Is there any properties or laws for this?
     
  18. Jul 10, 2013 #17
    There are a few ways you can think about [tex] \lim_{x \rightarrow 0} \frac{sin(2x)}{x} [/tex]. You can use the double angle formula for sine to write [itex] sin(2x)=2sin(x)cos(x) [/itex], which makes the limit clear, but this doesn't generalize very nicely to [tex] \lim_{x \rightarrow 0} \frac{sin(kx)}{x} [/tex] where [itex] k [/itex] is some other constant.

    Instead, you can write [tex] \lim_{x \to 0} \frac{sin(2x)}{x}=2 \lim_{x \to 0} \frac{sin(2x)}{2x}=2 \lim_{2x \to 0} \frac{sin(2x)}{2x}=2[/tex]

    Or, you can make the substitution [itex] u=2x [/itex], which turns the limit into [tex] \lim_{u \rightarrow 0} \frac{sin(u)}{(u/2)}=2 [/tex]. This suggests that a better original substitution would have been [itex] x=\frac{2}{t} [/itex]

    Of course, you could have just used L'Hopital's Rule on sin(2x)/x but there is no reason to rely on derivatives for this limit.

    tldr: [itex] \lim_{x \to 0} \frac{sin(kx)}{x}=k [/itex]
     
  19. Jul 10, 2013 #18
    Good explanation! Thank you very much!
     
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