Here is the original problem

So you do

whenever

Then you need to solve for x+1 to move on to the next step, but I don't know how...(Worried)

The book says you should get ... but I can't figure out how to get the absolute value.

- July 9th 2009, 12:42 PMChokfullLimits, absolute values, and inequalities with variables in the denominator
Here is the original problem

So you do

whenever

Then you need to solve for x+1 to move on to the next step, but I don't know how...(Worried)

The book says you should get ... but I can't figure out how to get the absolute value. - July 9th 2009, 04:58 PMhalbard
Better would be to write ( positive) when because .

- July 13th 2009, 11:42 AMChokfull
- July 13th 2009, 12:11 PMPlato
- July 13th 2009, 07:22 PMChokfull
- July 16th 2009, 01:13 PMChokfull
I'm trying to use the book's method to solve this problem.... The part I need help with is the absolute value in the denominator, I thought there was a rule for inequalities in which you can't multiply each side by the denominator, but I couldn't find it.

- July 16th 2009, 01:32 PMVonNemo19
http://www.mathhelpforum.com/math-he...8a78af75-1.gif

that for each number such that , a number for which .

You can see that the definition is satisfied.However, f(x) goes to NEGATIVE infinity. If you would like to prove it you could choose some arbitraily small number , solve the inequality , then use that value of x (to the left of -1) in the inequality (where a is -1), and show that however close x comes to a, f(x) will still be less than M. - July 16th 2009, 01:41 PMVonNemo19
- July 16th 2009, 03:45 PMdr.tea
as if whenever which yields

after solving the absolute value that so we can write which directly implies the result.