# What is the formal definition of a one sided limit?

I'm looking for the formal definition of $\displaystyle \lim_{x \to a^+}f(x) = L$ and $\displaystyle\lim_{x \to a^-}g(x) = M$

I took a guess at it intuitively, but I need to make sure this is correct:

$\displaystyle\lim_{x \to a^+} f(x) = L$ if and only if:

For any $\epsilon > 0$ there is a $\delta >0$ so that for any $x$, if $x-a<\delta$ then $f(x)-L < \epsilon$

$\displaystyle\lim_{x \to a^-}g(x) = M$ if and only if:

For any $\epsilon > 0$ there is a $\delta>0$ so that for any $x$, if $a-x<\delta$ then $M-g(x)<\epsilon$

• I found the answer. $\displaystyle \lim_{x \to a^+} f(x) = L$ if and only if: > For any $\epsilon >0$ there is a $\delta >0$ so that for any $x$, if $0 < x - a< \delta$ then $|f(x)-L|<\epsilon$ $\displaystyle \lim_{x \to a^-} f(x) = L$ if and only if: > For any $\epsilon >0$ there is a $\delta >0$ so that for any $x$, if $0 < a - x< \delta$ then $|f(x)-L|<\epsilon$ May 7, 2011 at 1:48
• You can post it as an answer, rather than a comment. May 7, 2011 at 6:03

$\displaystyle \lim_{x \to a^+} f(x) = L$ if and only if
For any $\epsilon>0$ there is a $\delta>0$ so that for any $x$, if $0 < x-a <\delta$ then $|f(x)-L| < \epsilon$.
$\displaystyle \lim_{x \to a^-} f(x) = L$ if and only if
For any $\epsilon>0$ there is a $\delta>0$ so that for any $x$, if $0 < a-x <\delta$ then $|f(x)-L| < \epsilon$.