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richyw

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

  • one year ago
  • one year ago

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

    • one year ago
  2. richyw
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    my textbook shows that\[\lim_{x\rightarrow 0}x\sin{\left(\frac{1}{x}\right)=0}\]because\[\left|x\sin{\left(\frac{1}{x}\right)}\right|\leq|x|\]

    • one year ago
  3. klimenkov
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    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 \(x=0\).

    • one year ago
  4. richyw
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    I don't understand how this shows that the limit is zero?

    • one year ago
  5. richyw
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    I mean it's clear that this inequality holds to me. but it's unclear to me how this finds the limit.

    • one year ago
  6. klimenkov
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    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!

    • one year ago
  7. richyw
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    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?

    • one year ago
  8. klimenkov
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    Yes. That is right.

    • one year ago
  9. richyw
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    thank you so much

    • one year ago
  10. klimenkov
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    You are welcome.

    • one year ago
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