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Infinite Limit help

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    Later in the problem I will use:

    [tex] \epsilon^{+} \;\;\;\;\;\;\;and\;\;\;\;\;\;\; \epsilon^{-} [/tex]

    to represent positive and negative infinitesimals, respectively.


    [tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2} [/tex]

    [tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}(1+2t^{-2})}}{4t+2} [/tex]

    [tex] \lim_{t \to -\infty} \frac {t\sqrt{1+2t^{-2}}}{4t+2} [/tex]

    [tex] \lim_{t \to -\infty} \frac {t\sqrt{1+2t^{-2}}}{t(4+2t^{-1})} [/tex]

    [tex] \lim_{t \to -\infty} \frac {\sqrt{1+2t^{-2}}}{4+2t^{-1}} [/tex]

    So at this point the limit is basically saying this:

    [tex] \frac {1 + \epsilon^{+}}{4+ \epsilon^{-}} [/tex]

    this is because the reciprocal of negative infinity squared is a positive infinitesimal and the reciprocal of negative infinity is a negative infinitesimal so the limit is:

    [tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2} = \frac {1}{4} [/tex]

    My book states it should be -1/4 but I do not see why. Your adding a positive infinitesimal to 1 and subtracting it from 4, but that doesn't make the numerator or denominator negative, so what I am I missing?
  2. jcsd
  3. May 18, 2010 #2
    Delete post
    Last edited: May 18, 2010
  4. May 18, 2010 #3
    \sqrt{t^{2}} = \left| t \right| = \left\{
    t & , & t \ge 0 \\
    -t & , & t < 0
    \end{array} \right.
  5. May 18, 2010 #4
    That too. So the above is a better answer. Multiply by |t|/|t| and look at the situation, ie. where the limit is going off to, if you have a problem like the one you have with a square root, etc. in order to see if you need a +t or -t.
  6. May 18, 2010 #5
    No! You only made a mistake in step 3 where you took [itex]t^{2}[/itex] from the square root. According to my formula, you should have taken something else. Everything else follows from there.
  7. May 18, 2010 #6
    You're right, I feel stupid now for not realizing my mistake from the beginning. They should have left the t^2 in there and multiplied the expression by |t|/|t| to take the t^2 away in the numerator and leave a -t to be multiplied through in the denominator and apply the limit to get the answer.
  8. May 18, 2010 #7
    Could you possibly explain what it is that I was suppose to take out? The (1+2t^{-2})? cause that would be the only part to take out of the radical and I don't see that getting me to the correct solution.


    I don't see why I would need to multiply t/t by -t/-t to get them to cancel just because it is -infinity, t/t = t*t^-1 = 1 inherently, why is the multiplication necessary?
  9. May 18, 2010 #8
    sorry, didnt notice the last post physicsman2,

    so should i not have even factored out the t^2 and instead multiplied by |t|/|t| which is effectively -t because the limit is - infinity? I still don't see how it is solved though...
  10. May 18, 2010 #9
    You solved it, though. When t < 0, |t|/t = -1, which is where the negative appears.
  11. May 18, 2010 #10
    oh, i am pretty sure i got it then
  12. May 18, 2010 #11
    Pretty much this.

    Sorry for the confusion earlier, I wasn't thinking all that well until I was corrected.
    Remember that |t| also equals sqrt(t^2). |t| can also equal either t or -t the way Dickfore showed. Since you want to take out a t^2 and cancel to apply the limit, you change |t| to sqrt(t^2) to bring it into the radical and divide. You would change the |t| in the denominator to -t because you're approaching -infinity, just as Tedjn showed. Divide and apply the limit, and you get -1/4
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