Suppose we want to prove $\lim_{x \to 2} g(x) = 4$ and $g(x) = x^2$
So we need to show for all $\epsilon > 0 \space \exists \space \delta>0$ such that
$0<|x-2|<\delta \implies |g(x) – 4| < \epsilon$ to complete this proof.
I am having difficulty in following the author's solution.
He says we can make $|x-2|$ as small as we like (makes sense, since $|x-2|<\delta$ for some positive $\delta$)
But we need an upper bound on $|x+2|$. If this term doesn't have an upper bound, then we don't know how to pick an appropriate $\delta$ so this also makes sense.
Now at this point onwards I am failing to understand what is happening.
Author says the $\delta$ neighbourhood is centred at $2$ (okay, since $2$ is the limit point of the domain of $g$), but that is must have a radius no bigger than $\delta = 1$.
Why is this the case?
Further more, we get the upper bound $|x+2| \leq |3+2| = 5$
Where does this last line come from?
If we let $\delta = 1$ (why?), this implies $|x-2| < 1$ $\implies$$2-1 < x < 2+1$
So then if $x<3$ and we seek an upper-bound for $|x+2| $, shouldn't it be
$|x+2| < 3+2=5$ as opposed to $|x+2| \leq 5$ ?
Best Answer
The limit game involves somebody giving you an epsilon, and then you get to choose your delta to try to satisfy the conditions. So, you get to choose to have $\delta \leq 1.$ And that is what the author has done. No matter what $\epsilon$ is served up, you decide to never pick a $\delta >1.$
You're right that the inequality can be made strict, but I expect that they're just using it justify a $\delta = \epsilon/5$ choice, and in that context it doesn't particularly matter. If $|x+2|< 5$ then it is surely the case that $|x+2|\leq 5.$