# Epsilon-delta inequality help

1. Jan 28, 2007

### Turtle1991

1. The problem statement, all variables and given/known data

Show that, if $$0<\delta<1$$ and $$|x-3|<\delta$$, then $$|x^{2}-9|<7\delta$$

2. Relevant equations

3. The attempt at a solution

I know that I need to transform one of the inequalities into the form of the other to prove it, but I don't see how. I can plug in values and of course it works since the limit of x^2 as x approaches 3 is 9, but I don't see how to show it algebraically. Please help

Last edited: Jan 28, 2007
2. Jan 28, 2007

### quasar987

|x²-9| = |x+3||x-3|<|x+3|d

From there, how can you use the triangle inequality and the hypothesis of |x-3|<d and 0<d<1 to get the result?

3. Jan 28, 2007

### HallsofIvy

Note that if |x-3|< $\delta$ and and $\delta$< 1, then |x-3|< 1 so -1< x-3< 1. What does that tell you about x+ 3 and |x+3|?