Definition of Limits

jikiami · #1 $Old$ September 18th, 2007, 06:48 PM

So i know the definition (the epsilon, delta one)... i know how to use, but when i'm given an exercise which i have to factorice (sp?) like for example (x^2 - 9) which turns to |(x+3)(x-3)| then i have to do something else, and i dont really understand that part.

Let's take this one

lim as x-->2 (x^2 - 1) = 3

Epsilon > 0 then there's a delta > 0 such that

0 < |x - 2| < delta, then |(x^2 - 1 - 3| < epsilon

0 < |x - 2| < delta, then |(x^2 - 4| < epsilon

0 < |x - 2| < delta, then |x+2| |x-2| < epsilon

Now the next step i know i have to pick a number, then do something i dont really understand.

2. Also i'm having trouble with this one on the same subject

f(x)=x^2 + 10x + 2, using the same delta epsilon argument to show that lim as x-->3 f(x)=f(3)

Epsilon > 0 then there's a delta > 0 such that

0 < |x - 3| < delta, then |x^2 + 10x + 2| < epsilon

I cant even get the function to look like |x - 3| :/

ThePerfectHacker · #2 $Old$ September 18th, 2007, 06:57 PM

Quote:

Originally Posted by jikiami $View Post$

So i know the definition (the epsilon, delta one)... i know how to use, but when i'm given an exercise which i have to factorice (sp?) like for example (x^2 - 9) which turns to |(x+3)(x-3)| then i have to do something else, and i dont really understand that part.

Let's take this one

lim as x-->2 (x^2 - 1) = 3

For $\epsilon > 0$ choose $\delta = \min \left\{ 1,\frac{\epsilon}{5} \right\}$ .

Thus, if $0<|x-2|<\delta$ that means $|x| \leq |x-2|+|2| < \delta + 2$ thus, $|x+2|\leq |x|+|2| < \delta + 4 \leq 5$ .
Thus,
$|x^2-1-3|=|x^2-4| = |x-2||x+2| < 5\delta \leq \epsilon$ .
Q.E.D.

jikiami · #3 $Old$ September 18th, 2007, 07:06 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

For $\epsilon > 0$ choose $\delta = \min \left\{ 1,\frac{\epsilon}{5} \right\}$ .

Thus, if $0<|x-2|<\delta$ that means $|x| \leq |x-2|+|2| < \delta + 2$ thus, $|x+2|\leq |x|+|2| < \delta + 4 \leq 5$ .
Thus,
$|x^2-1-3|=|x^2-4| = |x-2||x+2| < 5\delta \leq \epsilon$ .
Q.E.D.

Maybe you could explain me that as you would with someone living in the 3rd world? You did as to show the answer is right, but i just cant see it

My doubts... i dont know why you picked the min. you did, where does the 4 and 5 come from... i just dont get it

sorry