I'm a student taking a real analysis course and I have been asked the above question. I am new to but slowly figuring out the methods I must take to prove a limit from first principles but I am having trouble in the latter half of the method. So any tips or nudges in the right direction would be much appreciated!
Firstly I fix $\epsilon > 0$ and find $\delta > 0$ such that.
$$0 < |x- 2| < \delta \implies \bigg| \frac{x}{1+x} – \frac{2}{3} \bigg|< \epsilon $$
$$ \impliedby \bigg| \frac{x-2}{3(x+1)} \bigg| < \epsilon$$
$$\impliedby \frac{|x-2|}{3|x+1|} < \epsilon$$
I'm having trouble with the next part. I understand you need to pick a value, $n= something$ such that $|x-2| \leq n$, and $\frac{1}{3|x+1|} \geq n $. From here I know you have to pick an $\epsilon$ so. $\delta \leq min(n, ?)$ has the required property.
Sorry for the confusing question. I appreciate your time!
Best Answer
Let $|x-2|<1$, then
$-1<x-2<1$, or $2<x+1<4$;
Let $\epsilon >0$ be given.
Choose $\delta =\min (1, 6\epsilon);$
Then
|$x-2|<\delta$ implies
$|\frac{x-2}{3(x+1)}|< \frac{\delta}{3\cdot 2} \le \epsilon.$