# Precise definition of limits

• Sep 3rd 2007, 06:22 AM
fruitcakelover
Precise definition of limits
Does anyone knows how to show:

limit of f(x), x approaches a, equals L

if and only if

limit of f(x), x approaches a from the right = to limit of f(x), x approaches a from the left = L

Using the precise definition of limit.

• Sep 3rd 2007, 09:09 AM
ThePerfectHacker
Quote:

Originally Posted by fruitcakelover
Does anyone knows how to show:

limit of f(x), x approaches a, equals L

if and only if

limit of f(x), x approaches a from the right = to limit of f(x), x approaches a from the left = L

Using the precise definition of limit.

Let $\displaystyle f$ be defined on an open interval $\displaystyle I$ containing $\displaystyle a$ except possibly at $\displaystyle a$. With domain $\displaystyle D$.

Let, $\displaystyle \lim_{x\to a}f(x) = L$.
That means,
$\displaystyle 0 < |x-a| < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

We want to show,
$\displaystyle \lim_{x\to a^+}f(x) = L \mbox{ and }\lim_{x\to a^-}f(x) = L$.
Meaning,
$\displaystyle 0 < a - x < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.
$\displaystyle 0 < x - a < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

Do you see? Note that $\displaystyle |x-a|$ can either be $\displaystyle x-a \mbox{ or }a-x$ depending whether $\displaystyle x$ is larger than $\displaystyle a$ or not. But the point is that bpth of these quantieis $\displaystyle x-a, \ a-x$ are containted in the meaning of $\displaystyle 0<|x-a|< \delta$.
• Sep 3rd 2007, 03:24 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
Let $\displaystyle f$ be defined on an open interval $\displaystyle I$ containing $\displaystyle a$ except possibly at $\displaystyle a$. With domain $\displaystyle D$.

Let, $\displaystyle \lim_{x\to a}f(x) = L$.
That means,
$\displaystyle 0 < |x-a| < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

We want to show,
$\displaystyle \lim_{x\to a^+}f(x) = L \mbox{ and }\lim_{x\to a^-}f(x) = L$.
Meaning,
$\displaystyle 0 < a - x < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.
$\displaystyle 0 < x - a < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.

Do you see? Note that $\displaystyle |x-a|$ can either be $\displaystyle x-a \mbox{ or }a-x$ depending whether $\displaystyle x$ is larger than $\displaystyle a$ or not. But the point is that bpth of these quantieis $\displaystyle x-a, \ a-x$ are containted in the meaning of $\displaystyle 0<|x-a|< \delta$.

you forgot the immortal words "let $\displaystyle \epsilon > 0$" ...i don't know why, but i love that line, i'll say it again, "let $\displaystyle \epsilon$ be greater than zero" ...saying it the long way is even nicer

otherwise, good stuff!
• Sep 4th 2007, 01:29 AM
fruitcakelover
Thanks! I get it!