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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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And what have you tried so far? (If this is, by any chance, homework, please tag it with the [homework] tag). –  Arturo Magidin Feb 17 '11 at 2:58
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This is homework. just tagged it. I'm completely lost otherwise I would have tried it. –  morcutt Feb 17 '11 at 3:26
    
Thank you!. Some general advice: please try to put your entire question in the body (you'll notice I edited it to include what you needed to do in the body); don't rely on the subject for content, it makes life harder for your readers. Second, if you are not completely lost, then always write down what you've done so far. If you are completely lost, then say so too. It helps potential repliers know where to start or where to stop. –  Arturo Magidin Feb 17 '11 at 3:29
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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. –  Arturo Magidin Feb 17 '11 at 3:32

2 Answers 2

up vote 10 down vote accepted

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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I think something went terribly wrong after the displayed formula or I fail to understand what you're trying to say. –  t.b. Feb 17 '11 at 3:26
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I don't think your sarcastic condescension helps anyone. –  Ben Alpert Feb 17 '11 at 3:33
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In Arturo's defense, you can look over his past history of answering questions and see that he is always polite, welcoming, and quite helpful. While I can understand why some people read his tone in a negative way, I suspect he intended it light jest rather than condescension. –  Jason DeVito Feb 17 '11 at 3:46
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@All: Please, guys; yes, I meant it as funny, but I can definitely see Ben's point, which is why I took it out. –  Arturo Magidin Feb 17 '11 at 4:05
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@Arturo yes. It was very helpful. I apologize for all of the confusion. Thank you for your help. –  morcutt Feb 17 '11 at 4:26

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