need help with proofing limit (delta-epsilon)

[hyourinn]

i'm trying to prove that
the limit of (x+3) / (x²-3) is zero as x approaching ∞

so back to definition of limit

|f(x)-L|<E

|(x+3) / (x²-3) - 0|<E


(x+3) / (x²-3) <E ,and since the left side quite complicated i've come up with simple term that always bigger than (x+3) / (x²-3)
and that is (2x) / (x²/1.5) or 3/x . this, as long i take x>3 this should maintain :
(x+3) / (x²-3) <(3/x) inequality true.

so in first equality i take M=3 and it reflect x>M

then for second inequality:
(3/x)<E so x>(3/E) in other words in this inequality M= (3/E) and i know that both M is necessary to maintain inequality between

(x+3)/(x²-3) < (3/x) < E but in reality how both M affect me to choose x ? i mean in first inequality is demand me with x>3. so should i make second M follow this?

i mean second inequality (3/x) < E

x>3/E should i restrict E with 1>E so i can choose x>3 too ?

i'm sorry if it little hard to understand, please say something if you found my words was confusing , thanks

 

[pka]

prove that the limit of (x+3) / (x²-3) is zero as x approaching ∞
so back to definition of limit |f(x)-L|<E , |(x+3) / (x²-3) - 0|<E
then for second inequality: i mean second inequality (3/x) < E

It seems that you have all that you need done.
If $\displaystyle x>3$ then $\displaystyle \left|\frac{x+3}{x^2-3}\right|=\frac{x+3}{x^2-3}<\frac{3}{x}$. (***)
If you can establish (***) is true then you are there.
As $\displaystyle x\to \infty$ then $\displaystyle x>3$. Now if $\displaystyle \varepsilon>0$ so is $\displaystyle \frac{\varepsilon}{3}>0$.
Moreover, $\displaystyle (\exists M\in\mathbb{N}^+)\left[\frac{1}{M}<\frac{\varepsilon}{3}\right]$
Can you finish? HINT let $\displaystyle x>M$.

 

[hyourinn]

It seems that you have all that you need done.
If $\displaystyle x>3$ then $\displaystyle \left|\frac{x+3}{x^2-3}\right|=\frac{x+3}{x^2-3}<\frac{3}{x}$. (***)
If you can establish (***) is true then you are there.
As $\displaystyle x\to \infty$ then $\displaystyle x>3$. Now if $\displaystyle \varepsilon>0$ so is $\displaystyle \frac{\varepsilon}{3}>0$.
Moreover, $\displaystyle (\exists M\in\mathbb{N}^+)\left[\frac{1}{M}<\frac{\varepsilon}{3}\right]$
Can you finish? HINT let $\displaystyle x>M$.

first i'm very sorry for late reply, and i really grateful that you help me even though maybe it little hard to see my question ( i'm confuse how to use latex) , in this part that you wrote

$\displaystyle (\exists M\in\mathbb{N}^+)\left[\frac{1}{M}<\frac{\varepsilon}{3}\right]$

i should already solve that with x>(3/E) please check my first question on lower part, so back in my question in order to hold both inequality true i need keep
first x<3 and x<(3/E) so do i need restrict epsilon to 1 ? because if i try use more than 1 x will end up less than 3 and it will crumble the first inequality (when i simplify the main function)

 

[pka]

first i'm very sorry for late reply, and i really grateful that you help me even though maybe it little hard to see my question ( i'm confuse how to use latex) , in this part that you wrote
$\displaystyle (\exists M\in\mathbb{N}^+)\left[\frac{1}{M}<\frac{\varepsilon}{3}\right]$
i should already solve that with x>(3/E) please check my first question on lower part, so back in my question in order to hold both inequality true i need keep
first x<3 and x<(3/E) so do i need restrict epsilon to 1 ? because if i try use more than 1 x will end up less than 3 and it will crumble the first inequality

What does $\displaystyle x<3$ have anything to do with this? NOTHING
On the other hand, if $\displaystyle x\to\infty$ then it must be true that $\displaystyle x>3$ and also
$\displaystyle \left|\frac{x+3}{x^2-3}-0\right|<\frac{3}{x}$ can you agree with that. If not stop and find a live tutor.
Now here is a basic fact. For $\displaystyle (\forall \varepsilon>0)\left[\exists n\in\mathbb{N}^+\right]\text{ such that }\frac{1}{n}<\varepsilon$
If you put all of that together, you have a proof.
If you still do not know how to finish, then you definitely need to hire a personal tutor.

This is a free HELP service. But we do not offer tutorials. We only try to help you with what you misunderstand.