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Homework Help: Limit of hyperbolic cosine(1/x)

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Limit of x2cosh(1/x) as x approaches 0

    2. Relevant equations

    3. The attempt at a solution
    I did some algebra,made into the form infinity over infinity and used hopital's rule. I got limit x2e^(1/x).
    Then I again made into the form infinity over infinity(by making it 1/x2).
    Then used hopital's rule again and ended up getting limit of e^(1/x) as x approaches 0.

    Can I assume that it is infinity? But when I draw the graph of e^(1/x),using my graphic calculator, it shows the right and left limits are different and therefore the limit does not exist.

    Sorry for not showing my steps as I cannot find the limit and other symbols on latex.Thank you for reading!
  2. jcsd
  3. Aug 23, 2009 #2


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    Science Advisor

    I would be inclined to substitute u for 1/x so that [itex]x^2= 1/u^2[/itex] and that limit becomes
    [tex]\lim_{u\rightarrow \infty}\frac{e^u}{u^2}[/tex]
    Now using L'Hopital's rule (twice) will give you the answer more simply.
  4. Aug 23, 2009 #3
    Oh okay thank you.Btw Can you please explain to me why when the graph is drawn, there doesn't seem to be a limit as the right and left hand limits are different? Thanks again.
  5. Aug 23, 2009 #4


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    Homework Helper

    It is an even function therefore it is symmetric with respect to the y-axis. If what you say is true then the function would not be symmetric with respect to the y-axis hence not even. Perhaps you're plotting your function in a weird interval or are plotting the wrong function?
  6. Aug 23, 2009 #5
    Sorry I was ambiguous there.I am talking about the function e^(1/x).It does not seem to have a limit when I draw it on the graphing calculator.
  7. Aug 23, 2009 #6


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    Ah I missed that part. However using HallsofIvy's approach you won't have to deal with that limit. Instead you deal with the limit [itex]\lim_{u \to \infty} \frac{e^u}{u^2}[/itex], which is defined. Alternatively you could use the Taylor expansion of the cosh function to see that it diverges.
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