- #1

- 57

- 0

## Main Question or Discussion Point

As the thread title. The question is actually:

"Given p > 0, find d so that |x-0| < d implies |f(x)-1| < p and hence deduce that f(x) approaches 1 as x approaches 0."

My problem is that, when x is near the point x=0, x can be positive and negative, I don't know how to get my delta value because |f(x) -1| = |-x/(1+x)| can't be got into the |x-0| form without worrying about the signs of everything. It's confusing.

Any help?

"Given p > 0, find d so that |x-0| < d implies |f(x)-1| < p and hence deduce that f(x) approaches 1 as x approaches 0."

My problem is that, when x is near the point x=0, x can be positive and negative, I don't know how to get my delta value because |f(x) -1| = |-x/(1+x)| can't be got into the |x-0| form without worrying about the signs of everything. It's confusing.

Any help?