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Calculate the following limits Part 1

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]lim \underset{x \rightarrow }{\infty} {\frac{1+t^2}{1-t^2}}i+tan^{-1}(t)j+\frac{1-e^{-2t}}{t}k[/itex]

    First term = [itex]=\frac{1/t^2+1}{1/t^2-1}=-1[/itex]

    How do I show the second term and also the e^(-2t) term?

    Last edited by a moderator: Nov 10, 2011
  2. jcsd
  3. Nov 10, 2011 #2

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    Hi bugatti79! :smile:

    What is the range of [itex]\tan^{-1}[/itex]?
    If you don't know, you could for instance look it up in the definition of the tangent:

    Same for [itex]e^{-2t}[/itex].
    What is the smallest value that it can take (but just can't reach)?
  4. Nov 10, 2011 #3


    Staff: Mentor

    I don't understand the question. Is the limit as t --> infinity? You have x.
    If the limit is as t gets large, then yes.
    What do tan-1(t) and e-2t approach as t gets large?
  5. Nov 10, 2011 #4


    Staff: Mentor

    For this limit, use [ itex] \lim_{t \to \infty} ... [ /itex ]
  6. Nov 11, 2011 #5
    Ok. How about this one.

    I know the sin (t) as t tends to 0 = 0 but then how is the (sin (t))/t as t tends to 0 =1?

    Does y tend to 0 faster than sin(t) tends to 0?

    Similarly for cos(t) / t....? Is the converse true, ie cos (t) tends to 0 faster than y tends to 0.

    I.L.S, that is a good table in that wiki link. THanks.
  7. Nov 11, 2011 #6

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    You're moving on to a more complex limit.
    Does that mean you got the other limits?

    To find
    [tex]\lim\limits_{t \to 0} {\sin t \over t}[/tex]
    you should apply l'Hôpital's rule:
    The article gives your limit as an example.

    To answer your question, sin t and t approach 0 with equal speed.
  8. Nov 11, 2011 #7


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    Does cos(t) → 0 when t → 0 at all?
  9. Nov 11, 2011 #8
    Sorry, that should be cos(t) to 1 as t to 0
  10. Nov 11, 2011 #9
    Here is another example I am checking using different methods
    [itex]\frac{1-t^2+3t^5}{t^2+4}[/itex] as t tends to 0

    Method 1: Limit of the quotients is a quotient of the limits hence [itex]\frac{1-(0)^2+3(0)^5}{(0)^2+4}=\frac{1}{4}[/itex]

    Method 2: [itex]\frac{f(t)}{g(t)}=\frac{f'(t)}{g'(t)}[/itex] as t tends to 0 implies [itex]\frac{15t^4-2t}{2t}[/itex] as t tends to 0 gives [itex]\frac{0}{0}[/itex]....?

    Method 3: [itex]\frac{1/t^2-1/t^3+3}{1/t^3+4/t^5}=\frac{3}{0}[/itex]...?

    I have seen each of these methods work correctly in other problems....whats going on?
  11. Nov 11, 2011 #10

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    This is correct.

    L'Hôpital's rule is only applicable if nominator and denominator both approach zero, meaning you have the indeterminate form 0/0.
    This is not the case now.

    This form only works if t approaches infinity, which it doesn't in this case.
  12. Nov 11, 2011 #11
    Does this work for an indeterminate form of infinity over infinity also? Thanks
  13. Nov 11, 2011 #12

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    Actually, yes it works, as you can see in the wiki article.
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