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 |&lt; \epsilon implicates that x_{n} &gt; 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&lt; 0, then |x_n- L|= -(x_n- L)= L- x_n&lt; \epsilon so that -x_n&lt; -L+ \epsilon and, multiplying boty sides by -1, x_n&gt; L- \epsilon. Of course, if x_n- L\ge 0 then x_n\ge L and, since \epsilon&gt; 0, L&gt; L- \epsilon so x_n&gt; L-\epsilon.

In either case, x_n&gt; 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&lt; 0, then |x_n- L|= -(x_n- L)= L- x_n&lt; \epsilon so that -x_n&lt; -L+ \epsilon and, multiplying boty sides by -1, x_n&gt; L- \epsilon. Of course, if x_n- L\ge 0 then x_n\ge L and, since \epsilon&gt; 0, L&gt; L- \epsilon so x_n&gt; L-\epsilon.

In either case, x_n&gt; 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.