Originally Posted by

**mariogs379** Hey guys,

I'm trying to show that the limit as x-->1 of x^2 + 3x + 2 = 6.

We want to show that |x-1| < delta implies |(x^2 + 3x + 2) - 6| < epsilon.

|x^2 + 3x - 4| = |x-1| |x+4| < delta * |x+4|

So our delta = epsilon / |x+4|, which means that our delta changes as our x value changes, which we don't want. So we set delta = 1, which means:

|x-1| < 1

-1 < x-1 < 1

So 4 < x+4 < 6, which means that the biggest 1 / |x+4| can be is 1/4.

As a result, our delta = min{1, epsilon/4} (according to how Abbott does this in his book). Here's my question:

When we set delta = 1, we're confining the neighborhood around x=2 to (1, 3). But why does this mean that our delta = min{1, epsilon/4}? I'm a little bit confused about the jump in the last step to get to what delta has to be for a given epsilon > 0...

Thanks for the help guys,

Mariogs