Absolute Value & Limits

**Junesong** · February 22nd 2009, 09:27 PM

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 $f<img src=$ \rightarrow R" alt="f

\rightarrow R" /> has a limit at $x_{o}$ . Prove that $|f|<img src=$ \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, $\lim_{x \rightarrow x_{o}} |f(x)| = |L|$ .

**Moo** · February 23rd 2009, 10:28 AM

Originally Posted by Junesong

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

-----

Suppose that $f<img src=$ \rightarrow R" alt="f

\rightarrow R" /> has a limit at $x_{o}$ . Prove that $|f|<img src=$ \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, $\lim_{x \rightarrow x_{o}} |f(x)| = |L|$ .

It's okay

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