• Support PF! Buy your school textbooks, materials and every day products Here!

Limit of hyperbolic cosine(1/x)

  • #1

Homework Statement



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

Homework Equations





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!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955

Homework Statement



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

Homework Equations





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.
 
  • #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.
 
  • #4
Cyosis
Homework Helper
1,495
0
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?
 
  • #5
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.
 
  • #6
Cyosis
Homework Helper
1,495
0
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.
 

Related Threads on Limit of hyperbolic cosine(1/x)

  • Last Post
Replies
5
Views
5K
Replies
3
Views
9K
Replies
5
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
704
  • Last Post
Replies
9
Views
15K
  • Last Post
Replies
10
Views
951
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
4
Views
1K
Top