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.
 A: 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 $|x-6|$?
A: Since you said you're still lost after Arturo's post, I'll try to start earlier.
Q.  What do you mean intuitively by $\lim\limits_{x\rightarrow 6} \frac{x}{4} + 3$?
A.  Intuitively, you keep plugging in particular $x$ values really close to $6$ (but never actually plugging in $6$ - things like 5.99999 and 6.0000001 - into $\frac{x}{4}+3$ and you record the outputs.  Now, as you keep plugging in things closer and closer to $6$, you expect the outputs to hone in on one number.  The limit, then, is that one number.  By looking at the graph of $\frac{x}{4}+3$, you'd probably guess that the output is $\frac{6}{4} + 3 = \frac{9}{2}$.
Now, let me resay this answer in a way that will lead into the official math definition for a limit.
If I come along and say the limit is $\frac{9}{2}$, how would you test me?  Well, you could think to yourself "if the values are honing in on $\frac{9}{2}$, eventually they must get and stay within $.1$ of $\frac{9}{2}$, and so you challenge me by asking me to show that this is indeed the case.
Then I could respond by saying, "Once $x$ is within .01 of 6, then $\frac{x}{4} + 3$ will be within $.1$ of $\frac{9}{2}$.  For if $|x-6|<.01$, then $|\frac{x}{4}+3 - \frac{9}{2}| = |\frac{x}{4} - \frac{3}{2}| = |\frac{x-6}{4}| = \frac{|x-6|}{4} < \frac{.01}{4} < .1$."
If every time you come up with a tolerance (like $.1$), I can pass your test by making up a tolerance of my own ($.01$), then mathematically we'd say the limit is $\frac{9}{2}$.
Now, the official math definition is:
$\lim\limits_{x\rightarrow 6} \frac{x}{4} + 3 = \frac{9}{2}$ means for all $\epsilon > 0$, there is a $\delta > 0$ such that if $|x-6|<\delta$, then $|\frac{x}{4}+3 - \frac{9}{2}|< \epsilon$.
In our previous "conversation".  The $.1$ played the role of $\epsilon$ while the $.01$ played the role of $\delta$.
After reading (and  possibly rereading, and rerereading) all of the above, I'd encourage you to reread Arturo's response and see if you can turn what he said into a full fledged answer.
