Question

Another one of those limit problems, if it's approaching from the left or right it gives a different result:
Use the definitions of the hyperbolic functions to find the following limit: cosh (4)

And when I do lim x->0 (e^x + e^-x)/ (e^x-e^-x)
I get 2 results

Answer 1

Sorry its cothx

Answer 2

Interesting. You are correct, the limits from the left and right are not the same.

Answer 3

Hmm, I don't really know how to answer this one, if you were to guess, have any ideas on what would be my best bet?

Answer 4

myininaya, you were typing for a while, did ya have any ideas?

Answer 5

The limit does not exist, so I don't know what to tell ya :/

Answer 6

I thought that the rule is that if the limit from the right is different from the limit from the left, then the limit does not exist. Am I wrong?

Answer 7

No, that's correct.

Answer 8

So you think I should go for 0?

Answer 9

\[\lim_{x \rightarrow 0^+}\frac{e^x+e^{-x}}{e^x-e^{-x}} \cdot \frac{e^x}{e^x}\]
\[\lim_{x \rightarrow 0^+}\frac{e^{2x}+1}{e^{2x}-1}\]


\[\text{ let } u=e^{2x} ; \text{ as } x->0^+ , u->1\]

\[\lim_{u \rightarrow 1}\frac{u+1}{u-1}\]
We have vertical asymptote at u=1

Answer 10

lets make that as \[u->1^+\]

Answer 11

so we know that as u->1^+ then (u+1)/(u-1)->infinty

Answer 12

now if we look at as u->1^- then (u+1)/(u-1)-> -infinity

Answer 13

Yeah, and neither one of those are an option on the question

Answer 14

They both are but the point is the limit doesn't exist.

Answer 15

So do you have any ideas on what I should do?

Answer 16

the limit does not exist