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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.

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    So, please add that you are lost; it might help if you say what you *do* know: do you understand at least abstractly what the $\epsilon$-$\delta$ definition says? What you need to show? If not, say so. If you do, say so too. I'd already written an answer, but I am assuming you at least know what the definition is and what it means; if this is not the case, then it won't help, but I cannot know whether it is or not the case because you have not disclosed the status of your knowledge.2011-02-17

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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.

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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 $|x-6|$?

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    @Arturo yes. It was very helpful. I apologize for all of the confusion. Thank you for your help.2011-02-17