Limit of hyperbolic cosine(1/x)

[terminator88]

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!

 

[HallsofIvy]

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 $x^2= 1/u^2$ 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.

 

[terminator88]

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]

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]

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]

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.