When proving that
$\displaystyle \lim_{x\rightarrow p}(ax+b)=ap+b,a>0$
using the definition of the limit (or the -separated definition), find the largest possible value of when a=2 and =001
$\displaystyle \forall \epsilon > 0, \ \exists \delta > 0$ such that whenever $\displaystyle 0 < |x - p| < \delta$ it follows that $\displaystyle |(ax + b) - (ap + b)| < \epsilon$.
Looking at the latter expression and noting that a > 0:
$\displaystyle |(ax + b) - (ap + b)| = |ax - ap| = |a(x-p)| = a|x - p| < \epsilon$
So what $\displaystyle \delta$ should we choose?
What don't you understand?
Here's the formal definition of a limit:
$\displaystyle \forall \epsilon > 0, \ \exists \delta > 0$ such that whenever $\displaystyle 0 < |x - p| < \delta$ it follows that $\displaystyle |f(x) - L| < \epsilon$
Here: $\displaystyle f(x) = ax + b$ and $\displaystyle L = ap + b$.
With these epsilon-delta proofs, we play around with the expression $\displaystyle |f(x)-L| < \epsilon$ to get $\displaystyle |x-p| < m$ for some $\displaystyle m$.
Why do we do this? Because we know that $\displaystyle |x - p| < \delta$. So if we let $\displaystyle \delta = a$, then through all the algebra we did, it must necessarily follow that $\displaystyle |f(x) - L| < \epsilon$.
By definition, this proves that $\displaystyle \lim_{x \to p} f(x) = L$