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Homework Help: Limit understanding

  1. May 28, 2006 #1
    Okay so this is the problem
    Calculate this limit
    lim sin^2(3t)/t^2

    This is wat I did...

    I sed that it is also the lim sin3t*sin3t/t*t
    I know that sint is upper and lower bounded by -1 and 1
    Now this is where I just get completely stuffed... and confused!

    could someone just tell me or even show me how to think about doin the next step, and also tell me about the L'Hosp... rule... never been shown that and from other posts I've read it seems to make everythin a lot easier.
  2. jcsd
  3. May 28, 2006 #2
    for l'hopital's rule, you take the limit of the top derivative over the bottom derivative.
  4. May 29, 2006 #3
    For your question, what is
    [tex]\lim_{x\rightarrow 0}\frac{sin(x)}{x} ? [/tex]

    L'Hospital rule is applicable only in cases where the expression whose limit is to be taken gives an indeterminate form when evaluated directly(functional value 0/0 or inf./inf.)
  5. May 29, 2006 #4


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    Homework Helper

    And, in this case, it's the indeterminate form 0 / 0.
    Uhmm, you know the well-known limit:
    [tex]\lim_{t \rightarrow 0} \frac{\sin t}{t} = 1[/tex], right?
    That limit can be applied nicely here. I'll give you an example:
    [tex]\lim_{u \rightarrow 0} \frac{\sin (2u)}{u}[/tex]
    Since there is 2u in the numerator, while in the denominator there's only u, we'll try to multiply both numerator, and denominator by 2 to get:
    [tex]\lim_{u \rightarrow 0} \frac{\sin (2u)}{u} = \lim_{u \rightarrow 0} \frac{2 \sin (2u)}{2u} = 2 \lim_{u \rightarrow 0} \frac{\sin (2u)}{2u}[/tex]
    Now, as u tends to 0, 2u also tends to 0, right? If we let y = 2u, then as u -> 0, y also tends to 0, right?
    [tex]\lim_{u \rightarrow 0} \frac{\sin (2u)}{u} = 2 \lim_{y \rightarrow 0} \frac{\sin (y)}{y} = 2 \times 1 = 2[/tex].
    Is there anything unclear?
    Can you go from here? :)
    Last edited: May 29, 2006
  6. May 30, 2006 #5
    yeah cheers for that I should be able to figure it out from there
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