# Thread: another attempt to understand min/max value of lambda

1. ## another attempt to understand min/max value of lambda

i read and translated this "article" from a book that shows a new concept to me.
infinity small and infinitely large function.

can you explain to me using those terms

http://img262.imageshack.us/img262/1610/97335712sc3.gif

why in certain case we use maximum as the value for lambda and why
in another we use minimum as the value for lambda
like in here:

2. Originally Posted by transgalactic
i read and translated this "article" from a book that shows a new concept to me.
infinity small and infinitely large function.

can you explain to me using those terms

http://img262.imageshack.us/img262/1610/97335712sc3.gif

why in certain case we use maximum as the value for lambda and why
in another we use minimum as the value for lambda
like in here:
First, this is lambda: $\displaystyle \lambda$. And this is delta: $\displaystyle \delta$ .

As for your question: in this kind of proof, we need $\displaystyle \delta$ to be small, and sometimes we need $\displaystyle \delta$ to be smaller than, for instance, [tex]\delta_1[/Math] and $\displaystyle \delta_2$. Then this is equivalent to $\displaystyle 0\leq \delta\leq\min(\delta_1,\delta_2)$.
This is just taking $\displaystyle \delta$ small enough so that $\displaystyle \delta\leq\delta_1$ and $\displaystyle \delta\leq\delta_2$. If you know that something is true if [tex]|x-5|\leq \delta_1[/Math] and $\displaystyle |x-5|\leq\delta_2$, then this is true if $\displaystyle |x-5|\leq \delta=\min(\delta_1,\delta_2)$.

3. Originally Posted by transgalactic
Because here, for all $\displaystyle x>a$, we have $\displaystyle |f(x)|< M$ or $\displaystyle |f(x)|<|A|+\varepsilon$ (depending whether $\displaystyle x<c$ or $\displaystyle x>c$). And this is equivalent to saying that, for all $\displaystyle x>a$, $\displaystyle |f(x)|< \max(M,|A|+\varepsilon)$.

In the previous case, we wanted $\displaystyle \delta\leq 1$ and $\displaystyle \delta\leq \frac{\varepsilon}{11}$, which is equivalent to $\displaystyle \delta\leq \min(1,\frac{\varepsilon}{11})$.

The difference comes from the or or and. It is very natural as well since $\displaystyle \delta$ has to be small (we want something to be true for all $\displaystyle x$ such that $\displaystyle |x-5|<\delta$), while $\displaystyle B$ has to be large (we want $\displaystyle |f(x)|<B$).

4. why they use the Weierstrass law only on the 0<x<c case

why we cant say
for x>c
there is M>|f(x)|

(like in the couchy theorim for bounded sequence)
??

5. Originally Posted by transgalactic
why they use the Weierstrass law only on the 0<x<c case

why we cant say
for x>c
there is M>|f(x)|

(like in the couchy theorim for bounded sequence)
??
Because the Weierstrass law (as far as I guess what it is) only tells that continuous functions are bounded on segments, i.e. closed bounded intervals $\displaystyle [a,b]$. The function $\displaystyle x\mapsto x$ is continuous and obviously not bounded on $\displaystyle [0,+\infty)$.

In your situation, the function admits a limit $\displaystyle A$ at $\displaystyle +\infty$. In a way, this amounts to writing "$\displaystyle f(+\infty)=A$" and saying that "$\displaystyle f$ is continuous on $\displaystyle [0,+\infty]$" in order to apply Weierstrass rule. This is not fully rigorous as such, but with more mathematical knowledge you will be able to have this make sense. For now, there's no shortcut: the limit shows that $\displaystyle f$ is bounded (by $\displaystyle A+1$ for instance) on some interval $\displaystyle [c,\infty)$, and it is also bounded on $\displaystyle [a,c]$ by Weierstrass, so that it is bounded on all $\displaystyle \mathbb{R}$ (by the maximum of the two previous upper bounds).
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