A community for students.
Here's the question you clicked on:
 0 viewing
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^xe^x)
I get 2 results
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^xe^x) I get 2 results

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Interesting. You are correct, the limits from the left and right are not the same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0myininaya, you were typing for a while, did ya have any ideas?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The limit does not exist, so I don't know what to tell ya :/

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So you think I should go for 0?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0^+}\frac{e^x+e^{x}}{e^xe^{x}} \cdot \frac{e^x}{e^x}\] \[\lim_{x \rightarrow 0^+}\frac{e^{2x}+1}{e^{2x}1}\] \[\text{ let } u=e^{2x} ; \text{ as } x>0^+ , u>1\] \[\lim_{u \rightarrow 1}\frac{u+1}{u1}\] We have vertical asymptote at u=1

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0lets make that as \[u>1^+\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0so we know that as u>1^+ then (u+1)/(u1)>infinty

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0now if we look at as u>1^ then (u+1)/(u1)> infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, and neither one of those are an option on the question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0They both are but the point is the limit doesn't exist.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So do you have any ideas on what I should do?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0the limit does not exist
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.