Proving the value of a limit using the $\epsilon$-$\delta$ definition I'm trying to solve the problem of showing that
$$\lim_{x\to6}\left(\frac{x}{4}+3\right) = \frac{9}{2}$$
using the $\epsilon$-$\delta$ definition of a limit.

Given an $\epsilon\gt 0$, you want to show that provided $x$ is close enough to $6$, without being equal to $6$, then $\frac{x}{4}+3$ will be $\epsilon$-close to $\frac{9}{2}$. Well, first thing is to figure out how close $\frac{x}{4}+3$ is to $\frac{9}{2}$:
$$\left|\left(\frac{x}{4}+3\right) - \frac{9}{2}\right| = \left|\frac{x}{4}+3-\frac{9}{2}\right| = \left|\frac{x + 12 - 18}{4}\right| = \frac{|x-6|}{4}.$$
So, how can you make sure that $\displaystyle \left|\left(\frac{x}{4}+3\right) - \frac{9}{2}\right|$ is smaller than $\epsilon$, by placing conditions on how close $x$ is to $6$, that is, on the value of $|x-6|$?
