Limit of hyperbolic cosine(1/x)

  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. HallsofIvy

    HallsofIvy 41,264
    Staff Emeritus
    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. 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. Cyosis

    Cyosis 1,495
    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. 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. Cyosis

    Cyosis 1,495
    Homework Helper

    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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