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I'm trying to prove $$\lim_{x\to\infty}\frac{x+3}{x^2-3}=0$$ using delta-epsilon.

In the definition of limit

$$|f(x)-L|\lt\epsilon$$

$$|\frac{x+3}{x^2-3}-0|\lt\epsilon$$

$$|\frac{x+3}{x^2-3}|\lt\epsilon$$

and since the left side quite complicated I've come up with simple term that always bigger than $$\frac{x+3}{x^2-3}$$ and that is $$\frac{2x}{x^2/1.5}$$ or $$\frac{3}{x}$$ as long I take $x>3$. so here I take $3$ as $M$ so it will turn $x>M$ into $x>3$

then for second inequality:

$$\frac{3}{x}\lt\epsilon$$

$$x\gt\frac{3}{\epsilon}$$

and here I take $\frac{3}{\epsilon}$ as $M$ and wrapping things up

$$\frac{x+3}{x^2-3}\lt\frac{3}{x}\lt\epsilon$$.

But in reality how both $M$ affect me to choose $x$ ?

I mean in first inequality is demand me with $x>3$. In other words if I choose $M$ that smaller than $3$, it will crumble the first inequality (when I try simplifying things). So if I look at second $M (=\frac{3}{\epsilon})$ should I have the limitation $$\epsilon\lt1$$ so that I only can get $x$ more than $3$ ?

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pikarin-g
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    You can always ask that $M$ be the largest of $3$ and whatever your calculation with $\epsilon$ requires, so that all requirements are met at the same time. Note that anything larger than something that works will also work. And good on you for noticing there is something to make sure of! – Arturo Magidin May 05 '19 at 04:07
  • If $x>3$ then $ 0<1+\frac {3}{x}<2$ and $x-\frac {3}{x}>x-1>x/2>0... $ So $x>3\implies 0< \frac {x+3}{x^2-3}=$ $\frac {1+\frac {3}{x}}{x-\frac {3}{x}}<$ $\frac {2}{x/2}=$ $\frac {4}{x}.$ – DanielWainfleet May 05 '19 at 10:02
  • For finding the limit of $F(x)$ as $x\to \infty,$ it doesn't matter what $F(x)$ is for any $x\le 3.$ – DanielWainfleet May 05 '19 at 10:11

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