# Thread: Precise Definition of A Limit Problem

1. ## Precise Definition of A Limit Problem

Stuck on this one problem in my math homework:
http://i.imgur.com/2uVLo.png

2. ## Re: Precise Definition of A Limit Problem

I know how to find the precise definition of a limit but the way this one is set up is weird to me.

3. ## Re: Precise Definition of A Limit Problem

we want to find a δ > 0 so that when:

|x - 1/2| < δ, then

|f(x) - L| < ε = c

now f(x) = mx + b, and L = m/2 + b, so we need to find such a δ that makes:

|mx + b - (m/2 + b)| < c.

note that:

|mx + b - (m/2 + b)| = |mx + b - m/2 - b| = |m(x - 1/2)| = |m||x - 1/2| = m|x - 1/2| (since m > 0).

so what we really want is a δ such that:

|x - 1/2| < δ implies: |x - 1/2| < c/m. any ideas?

4. ## Re: Precise Definition of A Limit Problem

I am lost :\. I tried to go through the usual steps of finding the precise def of a limit but it looks long and confusing. I basically got c>o < mx+b - [(m/2) +b] < c>0 but I have no idea what to do from there.

5. ## Re: Precise Definition of A Limit Problem

Originally Posted by Mikusboi
I am lost :\. I tried to go through the usual steps of finding the precise def of a limit but it looks long and confusing. I basically got c>o < mx+b - [(m/2) +b] < c>0 but I have no idea what to do from there.
Surely you see that $\left| {mx + b - \left( {\frac{m}{2} + b} \right)} \right| = \left| m \right|\left| {x - \frac{1}{2}} \right|~?$
Let $\delta=\frac{c}{|m|+1}>0$

If $\left| {x - \frac{1}{2}} \right|<\delta$ then $\left| {mx + b - \left( {\frac{m}{2} + b} \right)} \right|