Precise definition of limits

**fruitcakelover** · September 3rd 2007, 06:22 AM

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.

**ThePerfectHacker** · September 3rd 2007, 09:09 AM

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

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

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

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

**Jhevon** · September 3rd 2007, 03:24 PM

Originally Posted by ThePerfectHacker

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

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

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

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

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

otherwise, good stuff!

**fruitcakelover** · September 4th 2007, 01:29 AM

Thanks! I get it!

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