Just need to check the following, can anyone help?

I need to use the epsilon-delta definition of continuity to prove that

f(x) = x^2 - 4x

is continuuous at 3

This is what I have,

The domain of f is R(the set of real numbers)

Let epsilon>0 be given. We want to choose delta>0, in terms of epsilon, such that

|f(x)-f(3)|<epsilon, for all x with |x-3|< delta (1.0)

1. First we write

f(x)-f(3) = x^2 - 4x + 3 = (x-1)(x-3)

2. Next we obtain an upper bound for |x+3|, when x is near 3.

If |x-3|<or equal to 1, than x lies in the closed interval [1,4], so

|x+3|<or eqaul to|x|+3(Triangle Inequality)

<or equal to 4+3=7

3. Hence

|f(x)-f(3)|<or equal to 7 |x-3|, for |x-3|<or equal to 1.

So if |x-3|< delta, where 0<delta<or equal to 1, then

|f(x)-f(3)|,7delta

Now 7delta<or equal to epsilon if and only if delta < or equal to 1/7 epsilon. Thus if we choose delta = min {1,1/7epsilon}, then

|f(x)-f(3)| < 7delta<or equal to 7 times 1/7 epsilon = epsilon, for all x with |x-3|< delta,

which proves staement (1.0).

Thus f is continuous at the point 3.