Epsilon, delta condition for limits questions

[Asphyxiated]

Homework Statement

Ok so this may get a little drawn out here because my book only gives me one example and I guess I can't decipher its meaning. So here is the example they give:

For \;\; \lim_{x \to 2} x^{2} = 4

Find a \;\; \delta > 0 \;\; such that whenever

0 < |x-2|< \delta, \;\;\; |x^{2}-4| < \frac {1}{10}

By the Lemma, we must find \delta > 0 such that whenever

2-\delta < x < 2+ \delta \;\;\;\; and\;\;\;\; x\neq2

4-\frac{1}{10} < x^{2} <4+\frac{1}{10}

(note that the only thing that the lemma states that that sentence is referring to is defining within delta and strictly within delta)

Assume that
2-\delta < x \;\;\;and\;\;\; x < 2+\delta

As long as \;\;\;2-\delta\;\;\; and \;\;\;x\;\;\; are positive we may square both sides and get:

4-4\delta+\delta^{2} < x^{2} \;\;\;\; and \;\;\;\; x^{2} < 4+4\delta+\delta^{2}

4+(-4\delta+\delta^{2}) < x^{2} \;\;\;\; and \;\;\;\; x^{2} < 4+(4\delta+\delta^{2})

(So here is question one, why did they square the inequality? Is it because f(x)=x^2 or some other reason that I am not picking up? This is where I would better understand what was going on with more than example being given)

Now take a \;\; \delta \;\; small enough so that:

-\frac {1}{10} \leq -4\delta + \delta^{2} \;\;\;\; and \;\;\;\; 4\delta+\delta^{2} \leq \frac {1}{10}

(So question two, why did they take only the \; -4\delta+\delta^{2} \; etc out instead of the entire left/right side? Is this only applicable to this problem and the squaring or does something similar happen on other epsilon, delta condition problems?)

Use 1/50 for example delta value and thus:

0<|x-2|<\frac {1}{50}, \;\;\; |x^{2}-4| < \frac {1}{10}

________________________________________________________

So an actual problem given at the end of the chapter is this:

For:

\lim_{x \to 2} \frac {1}{x} = \frac {1}{2}

Find a \; \delta > 0 \; such that whenever:

0 < |x-2| < \delta, \;\;\; |\frac{1}{x}-\frac{1}{2}| < .01

and the lemma (referenced above) states that x is strictly within \; \delta \; of c if and only if:

c-\delta<x<c+\delta

and in this problem c = 2 so this must be true:

2-\delta<x<2+\delta

so at this point I could make x into f(x) by doing a reciprocal like:

\frac{1}{2-\delta}<\frac{1}{x}<\frac{1}{2+\delta}

and if that is correct the next step i think would be this:

-.01 \leq \frac {1}{2-\delta} \;\;\;\; and \;\;\;\; \frac{1}{2+\delta} \leq .01

and if that is all correct I would just have to find a delta that is true for both those inequalities. I will stop here though so that I can get a yes no to my method. If this is way off then I apologize.

Thanks for any help!

 

[snipez90]

There are some issues with your solution to the problem, but if I answer the questions you have about the example perhaps you can spot the mistake.

The whole idea is of course is to get from 0 < |x-a| < delta to |f(x) - L| < epsilon, and often this is done backwards, starting with the conclusion. In the example, a = 2 and L = 4. The answer to both of your questions is basically the question "how can we manipulate what we already have to get our conclusion?" For question 1, the reason we need to square the inequality is because we don't have an x^2 term to begin with. Our conclusion however is 4 - (1/10) < x^2 < 4 + (1/10), so we need to get an x^2 by itself at some point. Since we have 2 - d < x < 2 + d, the best thing to do is to square the terms on both sides of each inequality, assuming we can take delta to be small enough so that the terms of each side of the inequality are positive. This ensures that we do have an x^2 term.

For question 2, look at why the author put -4d + d^2 and 4d + d^2 in parantheses. Once we actually have an x^2 term, the conclusion requires us to have 4 - (1/10) < x^2 < 4 + (1/10). But note that upon squaring the inequalities, we also get the 4 that we need in our conclusion, plus some delta terms. But we still need x^2 > 4 - (1/10), when we have x^2 > 4 -4d + d^2 upon squaring (I'm working with the greater than part now for simplification). If we can make 4 -4d + d^2 > 4 - (1/10) then by the transitive property x^2 is also > 4 - (1/10), which is part of our conclusion. But requiring 4 -4d + d^2 > 4 - (1/10) is the same as requiring -4d + d^2 > -1/10, since we may subtract the 4 from both sides (note also the inequality does not have to be strict, as the author uses a weak inequality). This is why the author considers -4d + d^2, and requires it to be greater than or equal to -1/10.

It takes some practice to get used to these ideas, but the basic principle is to work backwards from the conclusion. If you ever get stuck, remind yourself to always keep the conclusion in mind, and allow the next step in the proof to get you closer to the conclusion.

 

[Asphyxiated]

Oh wow thanks Snipez90, your explanation made way more sense than the authors. I am pretty sure I understand what is going on here now. I will try to do the book problem i posted tomorrow and post what I get here to see if it is correct. Many many thanks though Snipez90.

 

[Asphyxiated]

Ok, sorry it took me a day longer than I said to post this but I think need a little more help here if you will Snipez90. So using the same problem as before:

\lim_{x \to 2} \frac {1}{x} = \frac {1}{2}

find \; \delta &gt; 0 \; such \; that \;\; 0&lt;|x-2|&lt;\delta, \;\; |\frac{1}{x}-\frac{1}{2}| &lt; 0.01

Ok so going by what you said before I am trying to get:

2-\delta&lt;x&lt;2+\delta

to

\frac {1}{2}-.01 &lt; \frac {1}{x} &lt; \frac {1}{2} +.01

right? right...

So I think that the easiest way to get there from here is to take a reciprocal of both sides like I did before such that:

\frac {1}{2-\delta} &lt; \frac {1}{x} &lt; \frac {1}{2+\delta}

which is the same as:

\frac {1}{2} - \delta^{-1} &lt; \frac {1}{x} &lt; \frac {1}{2} + \delta^{-1}

and then if:

\frac {1}{2} - .01 &lt; \frac {1}{x} \;\;\; and \;\;\; \frac {1}{2}-\delta^{-1} &lt; \frac {1}{x}

then

\frac {1}{2}-.01 \leq \frac {1}{2}-\delta^{-1}

which again follow the example is to say:

-.01 \leq -\delta^{-1}

and it would be the same for the otherside like:

.01 \leq \delta^{-1}

So as of this point I would say that if we were to make delta greater than or equal to 100 the equalities would work out, at least how I set them up. This is obviously not the case though as the book gives delta for this problem as:

\delta = 2 - (.51)^{-1} \;\;\; or \;\;\; ~ .03922

I really don't see how they got that for delta at all, please tell me what I did wrong here...

 

[vela]

\frac {1}{2-\delta} &lt; \frac {1}{x} &lt; \frac {1}{2+\delta}

which is the same as:

\frac {1}{2} - \delta^{-1} &lt; \frac {1}{x} &lt; \frac {1}{2} + \delta^{-1}

That's not correct.

I suggest you start with

\left| \frac{1}{x}-\frac{1}{2} \right |&lt;0.01

and solve for x. That's typically the approach you take when you have an arbitrary ε to find a δ that will work.

 

[Asphyxiated]

Ok Vela, here it is perhaps solved for x, (I haven't worked with inequalities in a long time so if it is somehow wrong please be kind)

\frac {1}{x}-\frac {1}{2} &lt; \frac {1}{100}

note that I substituted 1/100 for .01

\frac{1}{x}&lt; \frac{1}{100}+\frac{1}{2}

\frac{1}{x} &lt; \frac {51}{100}

then the reciprocal to solve for x is:

x &lt; \frac {100}{51} \;\; or \;\; \approx 1.961

So how exactly does that help me get anywhere? This step was not in the examples nor had anything to do with what Snipez90 was telling me unless I just terribly misunderstood him. I am now even more confused than when I started...

I understand that we don't 'do others homework for them' but this is not homework for me it is just something I am trying to learn by myself. Can someone please just show me how this is properly done with this problem and maybe I can apply it to another problem in this section? I just feel that I am not going to get anywhere here with small suggestions in the right direction because I really don't know what the right direction is!

Please understand...

 

[vela]

Now subtract 2 from both sides since you want to know how close x has to be near 2. You also want to do the lower limit, so you need to do the same thing starting from 1/x-1/2 > -1/100.

 

[Asphyxiated]

Ok so you mean like:

x-2 &lt; \frac {100}{51} -2

and for the lower limit:

\frac {1}{x}-\frac{1}{2}&gt; -\frac{1}{100}

\frac {1}{x} &gt; -\frac{1}{100}+\frac{1}{2}

\frac {1}{x} &gt; \frac {49}{100}

x &gt; \frac {100}{49}

So would you subtract 2 again from this limit? Like:

x -2 &gt; \frac {100}{49}-2

which is actually:

x-2 &gt; \frac {2}{49}

and the upper limit is:

x-2 &lt; -\frac {2}{51}

but once again Vela, I still don't know where you are going with this technique as this is not done in any of the examples and once again I am completely lost, where does delta come in at in your method? So far I have done nothing to find delta that I can see.

 

[Asphyxiated]

also Vela where does:

c-\delta &lt; x &lt; c+\delta

come in in your technique? Is it even used at all?

 

[vela]

You need to reverse the inequalities when you take the reciprocal. Once you do that and put everything together, you'll have

-\frac{2}{51} &lt; x-2 &lt; \frac{2}{49}

So for the ε inequality to be true, x must be within 2/51 on the left and 2/49 on the right. On the number line, it looks like

<-----------(----------2--------------)--------->
             <- 2/51 -> <--  2/49  -->

You want to choose δ so that when -δ < x-2 < δ, x will be in that interval. Specifically, you need 0 < δ ≤ min{2/51, 2/49}=2/51. The book chose δ=1/51, which meets the criteria, but you can use any δ between 0 and 2/51.

 

[Asphyxiated]

Ok well this perhaps is just the way that you/we have the inequalities set up but in the book delta is:

\delta = 2 - (.51)^{-1} \;\; or \;\; \approx .03922

and your delta of just 1/51 is .01961, is that due to you having it as:

-\delta &lt; x -2 &lt; \delta \;\;\; instead \; of\;\;\; 2-\delta&lt;x&lt;2+\delta

? Or is there something something wrong?

 

[vela]

Sorry, I messed up. The book used δ=2/51=min{2/51, 2/49}, which is what 2-(0.51)^-1 is equal to.

 

[vela]

Is that due to you having it as:

-\delta &lt; x -2 &lt; \delta \;\;\; instead \; of\;\;\; 2-\delta&lt;x&lt;2+\delta

? Or is there something something wrong?

Those two statements are equivalent. I prefer the former because it plainly says that x has to be within δ of 2.

If you prefer the latter form, once you get to

-\frac{2}{51} &lt; x-2 &lt; \frac{2}{49}

you can add the 2 back again to get

2-\frac{2}{51} &lt; x &lt; 2+\frac{2}{49}

to make it easier to compare. You can again read off what the upper limit on δ is.

 

[Asphyxiated]

Hey Vela, if possible could you take a look at the new delta, epsilon thread I started? Thanks either way