1. ## Providing Conditions

I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?

2. Originally Posted by frenchguy87
I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?
$\displaystyle \left|x^2-1\right|=\left|x-1\right||x+1|$...so... come on.

3. Originally Posted by frenchguy87
I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?
$\displaystyle |x^2- 1|= |x-1||x+1|$ to make that less than $\displaystyle \frac{1}{n}$, you need to have $\displaystyle |x- 1|< \frac{1}{n|x+1|}$. But we need a constant, not a function of x on the right. That is, we need $\displaystyle |x- 1|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\displaystyle \frac{1}{x+1}$. If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}. Then $\displaystyle |x-1|< \frac{1}{n|x+1|}< \frac{1}{2n}$. In order that both of those be true, you must have |x-1|< the smaller of 1 and $\displaystyle \frac{1}{2n}$. Assuming that n is a positive integer, of course, $\displaystyle \frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $\displaystyle |x-1|< \frac{1}{2n}$.

4. Thanks a lot! Very helpful

5. You actually kind of lose me here, why do you say
Originally Posted by HallsofIvy
If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}.

6. Originally Posted by HallsofIvy
$\displaystyle |x^2- 1|= |x-1||x+1|$ to make that less than $\displaystyle \frac{1}{n}$, you need to have $\displaystyle |x- 1|< \frac{1}{n|x+1|}$. But we need a constant, not a function of x on the right. That is, we need $\displaystyle |x- 1|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\displaystyle \frac{1}{x+1}$. If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}. Then $\displaystyle |x-1|< \frac{1}{n|x+1|}< \frac{1}{2n}$. In order that both of those be true, you must have |x-1|< the smaller of 1 and $\displaystyle \frac{1}{2n}$. Assuming that n is a positive integer, of course, $\displaystyle \frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $\displaystyle |x-1|< \frac{1}{2n}$.
Shouldn't we choose |x-1|= min{ 1,$\displaystyle \frac{1}{3n}$ ???

,by adding 2 to all sides of -1< x-1<1 give you 1<x+1< 3

7. Originally Posted by frenchguy87
I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?
.

$\displaystyle |x-1|<\frac{1}{3n}$=> |x-1|<1 => |x|-1<1 => |x|+1<3 => |x+1|< 3 => $\displaystyle |x-1||x+1|\leq 3\frac{1}{3n}=\frac{1}{n}$

8. Originally Posted by xalk
.

$\displaystyle |x-1|<\frac{1}{3n}$=> |x-1|<1 => |x|-1<1 => |x|+1<3 => |x+1|< 3 => $\displaystyle |x-1||x+1|\leq 3\frac{1}{3n}=\frac{1}{n}$
Why do you guys all say $\displaystyle |x-1|<1$, I can't figure out where that comes from...

9. Originally Posted by CrazyCat87
Why do you guys all say $\displaystyle |x-1|<1$, I can't figure out where that comes from...
The point is to get some kind of bound on x so we can bound x+1. Since we want |x-1| small, |x-1|< 1 is handy. You could use any positive number other than 1 on the right, but 1 is easiest.

10. Originally Posted by HallsofIvy
$\displaystyle |x^2- 1|= |x-1||x+1|$ to make that less than $\displaystyle \frac{1}{n}$, you need to have $\displaystyle |x- 1|< \frac{1}{n|x+1|}$. But we need a constant, not a function of x on the right. That is, we need $\displaystyle |x- 1|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\displaystyle \frac{1}{x+1}$. If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}. Then $\displaystyle |x-1|< \frac{1}{n|x+1|}< \frac{1}{2n}$. In order that both of those be true, you must have |x-1|< the smaller of 1 and $\displaystyle \frac{1}{2n}$. Assuming that n is a positive integer, of course, $\displaystyle \frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $\displaystyle |x-1|< \frac{1}{2n}$.
I can almost follow this except for this part,
If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part).

Shouldn't we have 1 < x+1 < 3, which would then give

|x+1|> 1 so $\displaystyle \frac{1}{|x+1|}< 1$. Then $\displaystyle |x-1|< \frac{1}{n|x+1|}< \frac{1}{n}$ ?

11. This has gone on too long.

Do you agree that $\displaystyle |x-1|<1$ implies that $\displaystyle |x+1|<3$?

If so, then let $\displaystyle |x-1|<\frac{1}{3n}<1$.

That gives $\displaystyle |x^2-1|={\color{red}|x-1|}{\color{blue}|x+1|}<{\color{red}\left(\frac{1}{ 3n}\right)}{\color{blue}(3)}=\frac{1}{n}$.

Now you have your condition: $\displaystyle \frac{1}{3n}$.