# Thread: Absolute Value & Limits

1. ## Absolute Value & Limits

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

-----

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

2. 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 \rightarrow R" alt="f \rightarrow R" /> has a limit at $x_{o}$. Prove that $|f| \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