1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A limit problem

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

    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


    User Avatar
    Science Advisor
    Gold Member
    2017 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
    Have you seen L'Hospitals rule?
  9. Jul 9, 2013 #8
    I still can't do it :(
  10. Jul 9, 2013 #9


    User Avatar
    Science Advisor
    Homework Helper

    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

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

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


    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


    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted