Absolute value removed from equation. Why?

[jens.w]

Homework Statement

Given that $\epsilon$> 0, why is it, that $\left | x_{n} -L\right |< \epsilon$ implicates that $x_{n} > L-\epsilon$ ?

Homework Equations

The Attempt at a Solution

 

[Hurkyl]

Try some examples to get a feel for it.

Since you're dealing with absolute values, you could always use the usual technique for dealing with them: split into two cases depending on whether the argument is positive or negative.

 

[mtayab1994]

Well if: l Xn-L l ≥0 then you can remove the the absolute value.

 

[Ray Vickson]

Well if: l Xn-L l ≥0 then you can remove the the absolute value.

No, you can't. We have |-2| > 0; would you say that -2 > 0? Besides, for ANY real number r we always have |r| ≥ 0.

RGV

 

[LCKurtz]

Homework Statement

Given that $\epsilon$> 0, why is it, that $\left | x_{n} -L\right |< \epsilon$ implicates that $x_{n} > L-\epsilon$ ?

Remember that an equation like $|x_n-L|<\epsilon$ can always be rewritten $-\epsilon < x_n-L < \epsilon$. Add $L$ to all three sides and notice you are only using half of the result.

 

[jens.w]

Try some examples to get a feel for it.

Since you're dealing with absolute values, you could always use the usual technique for dealing with them: split into two cases depending on whether the argument is positive or negative.

Yes, i thought of that at first, but it doesn't help in this case since $x_{n}-L$ can be either positive or negative at all points, or alternating, depending on what series $x_{n}$ is a part of. So based on that, i can't exclude one of the cases via the definition.

 

[Ray Vickson]

Yes, i thought of that at first, but it doesn't help in this case since $x_{n}-L$ can be either positive or negative at all points, or alternating, depending on what series $x_{n}$ is a part of. So based on that, i can't exclude one of the cases via the definition.

If x >= L then x > L-ε (ε>0). So, just look at the case where x < L (assuming |x-L| < ε).

RGV

 

[HallsofIvy]

Homework Statement

Given that $\epsilon$> 0, why is it, that $\left | x_{n} -L\right |< \epsilon$ implicates that $x_{n} > L-\epsilon$ ?

As I am sure you know, |x|= x if $x\ge 0$, and |x|= -x if x< 0. So if $x_n- L< 0$, then $|x_n- L|= -(x_n- L)= L- x_n< \epsilon$ so that $-x_n< -L+ \epsilon$ and, multiplying boty sides by -1, $x_n> L- \epsilon$. Of course, if $x_n- L\ge 0$ then $x_n\ge L$ and, since $\epsilon> 0$, $L> L- \epsilon$ so $x_n> L-\epsilon$.

In either case, $x_n> L- \epsilon$.

 

[Hurkyl]

Yes, i thought of that at first, but it doesn't help in this case since $x_{n}-L$ can be either positive or negative at all points, or alternating, depending on what series $x_{n}$ is a part of. So based on that, i can't exclude one of the cases via the definition.

It doesn't matter that x_n comes from a series; the question you asked is one about the property of the number x_n.

And besides, in a proof by cases you don't exclude a cases: you consider all of them. (in this case, only 2)

 

[jens.w]

If x >= L then x > L-ε (ε>0). So, just look at the case where x < L (assuming |x-L| < ε).

RGV

Yea that makes sense.

As I am sure you know, |x|= x if $x\ge 0$, and |x|= -x if x< 0. So if $x_n- L< 0$, then $|x_n- L|= -(x_n- L)= L- x_n< \epsilon$ so that $-x_n< -L+ \epsilon$ and, multiplying boty sides by -1, $x_n> L- \epsilon$. Of course, if $x_n- L\ge 0$ then $x_n\ge L$ and, since $\epsilon> 0$, $L> L- \epsilon$ so $x_n> L-\epsilon$.

In either case, $x_n> L- \epsilon$.

That also makes sense! And how creative too.

It doesn't matter that x_n comes from a series; the question you asked is one about the property of the number x_n.

And besides, in a proof by cases you don't exclude a cases: you consider all of them. (in this case, only 2)

Your right, i was confused.

I guess the problem i had (besides being stupid of course) is that i got hung up on the results from the definition. They seemed to conflict and give 2 different cases, but that wasnt the case, which i feel that HallsOfIvy showed me.

Thank you all.