# 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.

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Suppose that $\displaystyle f \rightarrow R$ has a limit at $\displaystyle x_{o}$. Prove that $\displaystyle |f| \rightarrow R$ has a limit at $\displaystyle x_{o}$ and that $\displaystyle \lim_{x \rightarrow x_{o}} |f(x)| = | \lim_{x \rightarrow x_{o}} f(x) |$.

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

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.

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Suppose that $\displaystyle f \rightarrow R$ has a limit at $\displaystyle x_{o}$. Prove that $\displaystyle |f| \rightarrow R$ has a limit at $\displaystyle x_{o}$ and that $\displaystyle \lim_{x \rightarrow x_{o}} |f(x)| = | \lim_{x \rightarrow x_{o}} f(x) |$.

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