Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

richyw

  • 3 years ago

show that f is continuous at every point, \(a\in\mathbb{R}\)

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

    \[f(x)=x\sin{\left(\frac{1}{x}\right)} \text{ if } x\neq0\]\[f(0)=0\]

  2. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my textbook shows that\[\lim_{x\rightarrow 0}x\sin{\left(\frac{1}{x}\right)=0}\]because\[\left|x\sin{\left(\frac{1}{x}\right)}\right|\leq|x|\]

  3. klimenkov
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The function is continious in point \(x_0\) if \(\lim \limits_{x\rightarrow x_0}f(x)=f(x_0)\). You need to show that the function is continious in the point \(x=0\).

  4. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't understand how this shows that the limit is zero?

  5. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I mean it's clear that this inequality holds to me. but it's unclear to me how this finds the limit.

  6. klimenkov
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Think about it. If \(\left|x\sin{\left(\frac{1}{x}\right)}\right|\leq|x|\) and \(x\rightarrow0\), so \(|x|\) is very little, and then \(\left|x\sin{\left(\frac{1}{x}\right)}\right| \) is little too!

  7. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah, perfect. I get it now. so now because the limits exist, and f(0) is defined, and the limit= f(0), that should be satisfactory to show that the function is cts right?

  8. klimenkov
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes. That is right.

  9. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you so much

  10. klimenkov
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You are welcome.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy