# Limit of hyperbolic cosine(1/x)

by terminator88
Tags: cosine1 or x, hyperbolic, limit
 P: 60 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!
Math
Emeritus
Thanks
PF Gold
P: 39,682
 Quote by terminator88 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!
I would be inclined to substitute u for 1/x so that $x^2= 1/u^2$ and that limit becomes
$$\lim_{u\rightarrow \infty}\frac{e^u}{u^2}$$
Now using L'Hopital's rule (twice) will give you the answer more simply.
 P: 60 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.
 HW Helper P: 1,495 Limit of hyperbolic cosine(1/x) 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?
P: 60
 Quote by Cyosis 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?
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.
 HW Helper P: 1,495 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 $\lim_{u \to \infty} \frac{e^u}{u^2}$, which is defined. Alternatively you could use the Taylor expansion of the cosh function to see that it diverges.

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