# Finding modulus of continuity

1. Apr 28, 2009

### JG89

Say f(x) = x^2 - 1 and I'm trying to prove that f is continuous, then I was told I CANNOT do this:

$$|x^2 - x_0^2| = |x-x_0||x+x_0| < \delta|x+x_0| = \epsilon$$

because then our epsilon is relying on an x value. I was told I could restrict the x values to a closed neighborhood about the point x = x_0, and let epsilon rely on the end points, but I cannot let it rely on any x values like in the example shown.

Is this correct? And if it is, why is it?

2. Apr 28, 2009

### HallsofIvy

Staff Emeritus
Yes, it's true. It's true because $\epsilon$ is supposed to be a constant and $\delta|x+ x_0|$ is not a constant. so "$\delta|x+ x_0|= \epsilon$ makes no sense.

What you can do is this: Since we only need look at x close to x0, assume |x- x0|< 1. The -1< x- x0< 1 so, adding 2x0 to each part, -1+ 2x0< x+ x_0< 1+ 2x0. The larger of those two numbers, in absolute value, is 1+ 2x0 so $\delta|x+ x_0|< \delta(1+ 2x_0$. We must choose $\delta$ such that $\delta(1+ 2x_0)< \epsilon$. Actually, because of the requirement that |x- x0|< 1, we must choose $\delta$ to be the smaller of that number and 1.