A community for students.
Here's the question you clicked on:
 0 viewing
owlet
 one year ago
Question below
owlet
 one year ago
Question below

This Question is Closed

owlet
 one year ago
Best ResponseYou've already chosen the best response.0@SolomonZelman @Loser66 @triciaal

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Have you looked at a graph of f(x)? Have you plugged in x=0? or x=4? Have you looked at limits as we approached these numbers?

owlet
 one year ago
Best ResponseYou've already chosen the best response.0we're not allowed to use graphing devices. the answer is, 0 is the V.A. but 4 is not a V.A. but one of the zeros in the denominator is 4, so it should be a V.A. this is the part i'm not sure

freckles
 one year ago
Best ResponseYou've already chosen the best response.3at x=4 you should get 0/0 when you plug in the 4 which means it could be a hole at x=4 at x=0 you should get 2/0 which tells us we have a va at x=0 if you ever get something/0 where the top something isn't 0 then yep it is a va but if you get 0/0 you have more work

owlet
 one year ago
Best ResponseYou've already chosen the best response.0how to prove it using limits?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3cot(4) clearly exists but the problem is the other function evaluated at x=4 \[\lim_{x \rightarrow 4} \frac{\sqrt{x^2+9}5}{x^2+5x+4} \cdot \frac{\sqrt{x^2+9}+5}{\sqrt{x^2+9}+5}\] try evaluating this limit notice I multiplied that one function by a fancy one

owlet
 one year ago
Best ResponseYou've already chosen the best response.0okay. I get that part now. thanks.. how about the 0?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you should see this limit exists which tells us we don't have a va

freckles
 one year ago
Best ResponseYou've already chosen the best response.3anyways if you plug in x=0 you get something/0 where the top something isn't 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.3this means you have a va at x=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.3consider evaluating this limit if you are not sure \[\lim_{x \rightarrow 0}\cot(x)\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I think if you just multiple the term as what freckles did, and then simplify, x4 no more in denominator. Hence, no need to take the lim

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0And that is enough to conclude that x =4 is not a V.A

freckles
 one year ago
Best ResponseYou've already chosen the best response.3(x+4) ? and x=4? @Loser66 :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.3by the way if a limit does not exist the answer doesn't always say the discontinuity is a vertical asymptote we just knew our function didn't have any jump discontinuities

freckles
 one year ago
Best ResponseYou've already chosen the best response.3more details: \[\lim_{x \rightarrow a^{\pm}} f(x)=\pm \infty \text{ then } va \text{ at } x=a \\\]
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.