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Limit of hyperbolic cosine(1/x)

by terminator88
Tags: cosine1 or x, hyperbolic, limit
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terminator88
#1
Aug23-09, 10:26 AM
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!
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HallsofIvy
#2
Aug23-09, 11:11 AM
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Quote Quote by terminator88 View Post
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 [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.
terminator88
#3
Aug23-09, 11:35 AM
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.

Cyosis
#4
Aug23-09, 11:41 AM
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?
terminator88
#5
Aug23-09, 11:50 AM
P: 60
Quote Quote by Cyosis View Post
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.
Cyosis
#6
Aug23-09, 12:06 PM
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 [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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