# Thread: help visualising a limit proof..

1. i tried to interpret it on a sketch

http://img181.imageshack.us/img181/6498/13549423mx3.gif

i cant see the animation of this proove

2. If you take the definition of the limit:
$\displaystyle \forall\mathcal{O}(L)\;\exists\mathcal{P}(a)\;\for all x\in D(f)\cap\mathcal{P}(a):f(x)\in\mathcal{O}(L)$
You may also understand this by delta-epsilon notation:
$\displaystyle |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$
where $\displaystyle \mathcal{O}(L):=(L-\varepsilon;L+\varepsilon)$ and $\displaystyle \mathcal{P}(a):=(a-\delta;a+\delta)\backslash\{a\}$
To prove that
$\displaystyle \lim_{x\to p}f(x)=p$
you need to use the definition:
$\displaystyle |x-p|<\delta\Rightarrow|f(x)-p|<\varepsilon$
where $\displaystyle f(x)=x$ so I can rewrite:
$\displaystyle |x-p|<\delta\Rightarrow|x-p|<\varepsilon$
Now I can choose $\displaystyle \delta=\varepsilon$. So I prooved, that for all epsilons exists such delta that $\displaystyle |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$

Another example: Proove that $\displaystyle \lim_{x\to 1}(2x+2)=4$.
Solution: Again, I have to use definition: $\displaystyle |x-1|<\delta\Rightarrow|(2x+2)-4|<\varepsilon$. Now I have to find such delta, that satisfies epsilon. It is easy:
$\displaystyle |(2x+2)-4|=|2x-2|=|2|\cdot\underbrace{|x-1|}_{\delta}<\varepsilon$
There's another trick:
$\displaystyle |x-1|<\frac{\varepsilon}{2}=\delta$
So now, if you tell me some epsilon, I am able to tell you delta which satisfies your epsilon. If you tell $\displaystyle \varepsilon=0.001$ my delta is $\displaystyle \delta=0.0005$

3. Originally Posted by lukaszh
If you take the definition of the limit:
$\displaystyle \forall\mathcal{O}(L)\;\exists\mathcal{P}(a)\;\for all x\in D(f)\cap\mathcal{P}(a):f(x)\in\mathcal{O}(L)$
You may also understand this by delta-epsilon notation:
$\displaystyle |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$
where $\displaystyle \mathcal{O}(L):=(L-\varepsilon;L+\varepsilon)$ and $\displaystyle \mathcal{P}(a):=(a-\delta;a+\delta)\backslash\{a\}$
To prove that
$\displaystyle \lim_{x\to p}f(x)=p$
you need to use the definition:
$\displaystyle |x-p|<\delta\Rightarrow|f(x)-p|<\varepsilon$
where $\displaystyle f(x)=x$ so I can rewrite:
$\displaystyle |x-p|<\delta\Rightarrow|x-p|<\varepsilon$
Now I can choose $\displaystyle \delta=\varepsilon$. So I prooved, that for all epsilons exists such delta that $\displaystyle |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$

Another example: Proove that $\displaystyle \lim_{x\to 1}(2x+2)=4$.
Solution: Again, I have to use definition: $\displaystyle |x-1|<\delta\Rightarrow|(2x+2)-4|<\varepsilon$. Now I have to find such delta, that satisfies epsilon. It is easy:
$\displaystyle |(2x+2)-4|=|2x-2|=|2|\cdot\underbrace{|x-1|}_{\delta}<\varepsilon$
There's another trick:
$\displaystyle |x-1|<\frac{\varepsilon}{2}=\delta$
So now, if you tell me some epsilon, I am able to tell you delta which satisfies your epsilon. If you tell $\displaystyle \varepsilon=0.001$ my delta is $\displaystyle \delta=0.0005$
Of course are you are correct ! But for the benefit of the poster you might want to mention that $\displaystyle \mathcal{P}(a)$ is a neighborhood of radius $\displaystyle \delta$ around $\displaystyle p$...things like that