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Math Help - Absolute Value & Limits

  1. #1
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    Absolute Value & Limits

    I'm sure this is a really simple problem, I just wanted to make sure that I'm doing this correctly.

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    Suppose that \rightarrow R" alt="f \rightarrow R" /> has a limit at x_{o}. Prove that \rightarrow R" alt="|f| \rightarrow R" /> has a limit at x_{o} and that \lim_{x \rightarrow x_{o}} |f(x)| = | \lim_{x \rightarrow x_{o}} f(x) |.

    Proof: Let L = \lim_{x \rightarrow x_{o}} f(x). Then, \forall \epsilon > 0, \exists \delta >0 such that, if 0<|x-x_{o}|< \delta, |f(x)-L|< \epsilon. By the Inverse Triangle Inequality, we have that ||f(x)|-|L|| \leq |f(x)-L| < \epsilon. Therefore, \lim_{x \rightarrow x_{o}} |f(x)| = |L|.
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    Quote Originally Posted by Junesong View Post
    I'm sure this is a really simple problem, I just wanted to make sure that I'm doing this correctly.

    -----

    Suppose that \rightarrow R" alt="f \rightarrow R" /> has a limit at x_{o}. Prove that \rightarrow R" alt="|f| \rightarrow R" /> has a limit at x_{o} and that \lim_{x \rightarrow x_{o}} |f(x)| = | \lim_{x \rightarrow x_{o}} f(x) |.

    Proof: Let L = \lim_{x \rightarrow x_{o}} f(x). Then, \forall \epsilon > 0, \exists \delta >0 such that, if 0<|x-x_{o}|< \delta, |f(x)-L|< \epsilon. By the Inverse Triangle Inequality, we have that ||f(x)|-|L|| \leq |f(x)-L| < \epsilon. Therefore, \lim_{x \rightarrow x_{o}} |f(x)| = |L|.
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