Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

3psilon

  • one year ago

Does this limit exists as x-> 1 ?

  • This Question is Closed
  1. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[f(x) = (x ^{2}-1) x \neq 1 \]\[f(x) = (4) x = 1 \]

  2. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(x) = 4 at x = 1

  3. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    It's a polynomial, meaning it is continuous for all real numbers meaning you can just plug it in to find the limit.

  4. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1361493037294:dw|

  5. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So it still exists?

  6. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Wait, so we have \[ f(x) = \begin{cases} x^2-1 & x\neq 1\\ \text{undefined} & x=1 \end{cases} \]Right?

  7. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes I just couldn't get it to look like that. Undefined should be 4 though .

  8. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So it's just \[ \lim_{x\to1}f(x) = (1)^2-1 \]

  9. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  10. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    We can get really really close to 1 and we'll approach 0

  11. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Do you understand the epsilon delta definition of limit?

  12. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    On the graph it's not continuous but does the limit still exist is what I'm asking

  13. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, it has just been a while since i've done limits

  14. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    When you want to know if a limit exists, just look at the right hand and left hand limits.

  15. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    If they meet, then the limit exists.

  16. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But @wio They don't meet at that point cause at X=1 the point is shifted up

  17. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What are you getting for the left and right limits?

  18. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't know what to do because the point isn't there. It's shifted up

  19. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well if you wanted to find the right hand limit, you're going to plug in things like .999 and when you wanted to find the left hand limit, you're going to plug in things like 1.0001

  20. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @wio this is not helping at all. Can you please just answer my question if it exists or not. Not trying to be rude but I'd appreciate it

  21. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Did you plug in numbers getting close to 1 on both sides? What are they approaching?

  22. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    4

  23. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    No, it is approaching 0.

  24. 3psilon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So the limit does exist as it approaches 1?

  25. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \((0.9999)^2-1 \approx 0\)

  26. wio
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah.

  27. TuringTest
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    actually\[f(x) = \begin{cases} x^2-1 & x\neq 1\\ 4 & x=1 \end{cases}\]but that is not important. if all we are asking is if the limit exists, then @wio is heading you in the right direction. only if they are asking if it is continuous does the other part matter.

  28. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find 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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.