anonymous 4 years ago 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

1. anonymous

Sorry its cothx

2. anonymous

3. anonymous

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

4. anonymous

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?

5. anonymous

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

6. anonymous

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

7. Mertsj

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?

8. anonymous

No, that's correct.

9. anonymous

So you think I should go for 0?

10. myininaya

$\lim_{x \rightarrow 0^+}\frac{e^x+e^{-x}}{e^x-e^{-x}} \cdot \frac{e^x}{e^x}$ $\lim_{x \rightarrow 0^+}\frac{e^{2x}+1}{e^{2x}-1}$ $\text{ let } u=e^{2x} ; \text{ as } x->0^+ , u->1$ $\lim_{u \rightarrow 1}\frac{u+1}{u-1}$ We have vertical asymptote at u=1

11. myininaya

lets make that as $u->1^+$

12. myininaya

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

13. myininaya

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. myininaya

the limit does not exist