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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 $M-g(x)<\epsilon$

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    $\begingroup$ 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 $|f(x)-L|<\epsilon$ $\endgroup$ May 7, 2011 at 1:48
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    $\begingroup$ You can post it as an answer, rather than a comment. $\endgroup$ May 7, 2011 at 6:03

1 Answer 1

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$\displaystyle \lim_{x \to a^+} f(x) = L$ 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 $|f(x)-L| < \epsilon$.

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