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