TITLE: Limit of a Function as x goes to Infinity: Proof Verification Using the Definition
QUESTION [0 upvotes]: I have written out a proof for this question, although it feels wrong.
$\lim _{x\to \infty }\left(\frac{x^2}{x^2+5x+5}\right)$
Take any Ɛ>0, I started by finding $|\frac{x^2}{x^2+5x+5}-1|$, which gives me $\frac{5|x+1|}{|x^2+5x+5|}$.
From here is where I am confused.
I restricted $d\leq2$, so $|x+1|\leq2$ and $-3\leq x\leq1$.
I factored $x^2+5x+5$ to get $(x+\frac{5+\sqrt{5}}{2})(x+\frac{5-\sqrt{5}}{2})$, and then using the above inequality for x I found that $|\frac{1}{(x+\frac{5+\sqrt{5}}{2})}|\leq\frac{2}{-1+\sqrt{5}}$ and $|\frac{1}{(x+\frac{5-\sqrt{5}}{2})}|\leq\frac{2}{1+\sqrt{5}}$.
Then: $\frac{5|x+1|}{|x^2+5x+5|}\leq 5|x+1|(\frac{2}{1+\sqrt{5}})(\frac{2}{-1+\sqrt{5}})=\frac{5}{2}|x+1|$.
So $\frac{5}{2}|x+1|\leqƐ$ iff $|x+1|\leq\frac{2Ɛ}{5}$.
Now choose d=max{2, $\frac{2Ɛ}{5}$}, given Ɛ>0.
This proof feels wrong, although I am confused as to how to better prove limits where x goes to infinity, as the text I am using as reference only shows a very simple example.
Any help would be greatly appreciated!

REPLY [1 votes]: Option:
$x>0:$
$f(x):=\dfrac{5(x+1)}{x^2+5x+5} <$
$\dfrac{5(x+1)}{(x+1)^2} = \dfrac{5}{x+1}<\dfrac{5}{x}$.
$\epsilon >0$ given.
Choose $M = 5/\epsilon$.
For $x > M:$
$f(x)< \dfrac{5}{x} <5/M =\epsilon$.
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