deltas for limits

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Oct 30th 2007, 04:36 PM
PvtBillPilgrim

deltas for limits

For a given epsilon e > 0, I need to find a delta d such that for all x, abs(x) < d.

I have these three:
abs(sqrt(x+4) - 2) < e

For this, I rationalized abs(sqrt(x+4) - 2) and got abs(x/(sqrt(x+4)+2), which I said is <= abs(x/2). Then I chose delta to be e/2. Is this ok?

The second one is abs(((1-x)^4) - 1) < e and the third one is abs([(x^2 - x + 1)/(x+1)] - 1) < e. I'm not really sure how to do these. Could someone show me how to find the appropriate d(e)? I'm somewhat lost at the moment and I have a whole bunch to solve.
Oct 30th 2007, 07:24 PM
ThePerfectHacker

Quote:

Originally Posted by PvtBillPilgrim

For a given epsilon e > 0, I need to find a delta d such that for all x, abs(x) < d.

I have these three:
abs(sqrt(x+4) - 2) < e

For this, I rationalized abs(sqrt(x+4) - 2) and got abs(x/(sqrt(x+4)+2), which I said is <= abs(x/2). Then I chose delta to be e/2. Is this ok?

The second one is abs(((1-x)^4) - 1) < e and the third one is abs([(x^2 - x + 1)/(x+1)] - 1) < e. I'm not really sure how to do these. Could someone show me how to find the appropriate d(e)? I'm somewhat lost at the moment and I have a whole bunch to solve.

$\left| \sqrt{x+4} - 2 \right| = \left| \frac{x}{\sqrt{x+4}+2} \right| = \frac{|x|}{\sqrt{x+4}+2}$ .

If $|x| < 1 \implies -1 < x < 1 \implies 3 < x+4 < 5 \implies \sqrt{3} < \sqrt{x+4} < \sqrt{5} \implies$ $\sqrt{3}+2 < \sqrt{x+4} + 2 < \sqrt{5} + 2$ .

Thus, if $0< \delta \leq 1$ then $|x| < \delta \implies \sqrt{x+4}+2 > \sqrt{5}+2$ .

This means,
$\frac{|x|}{\sqrt{x+4}+2} \leq \frac{|x|}{\sqrt{5}+2} < |x| < \delta$ .

Thus, for $\epsilon > 0$ choose $\delta = \min \{ \epsilon , 1 \}$ .
Oct 30th 2007, 07:39 PM
PvtBillPilgrim

Can you quickly show me how to do the other two? I have to do like ten and some of them look similar to these, so if I could see how they are done, I should be ok. Also, I don't really see where this is coming from exactly. We're learning about the delta-epsilon notation of limits of functions, however there's no accumulation point c here for 0 < abs(x-c) < d(e). So how exactly are you picking the deltas?
Oct 30th 2007, 07:48 PM
ThePerfectHacker

$|(x-1)^4 - 1| = |x^4 - 4x^3 + 6x^2 - 4x + 1 - 1| \leq |x|^4+4|x|^3+6|x|^2 + 4|x|$ .
If $0 < \delta \leq 1$ then, (given that $0<|x|<\delta$ ).
$|x|^4+4|x|^3 + 6|x|^2 + 4|x| \leq \delta^4 + 4\delta^3 + 6\delta^2 + 4\delta \leq \delta + 4\delta + 6\delta + 4\delta = 15\delta$ .
So choose,
$\delta = \min \left\{ \frac{\epsilon}{15} , 1 \right\}$ .
Oct 30th 2007, 08:15 PM
PvtBillPilgrim

Thanks for the help man.

For the last one, I get the absolute value of:
x^2 - 2x
--------, which is less than or equal to the absolute value of
x + 1

x^2 - 2x
--------, which equals the absolute value
x

x-2, which is less than or equal to

abs(x) + 2.

How do I then find the delta for this?