Thread: Formally proving limits with epsilon-delta proofs

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

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

Well actually we're not confining the neighbourhood around x = 2 (i.e. (1, 3) ), we're actually confining the neighbourhood around x = 1 (i.e. (0, 2) ).

What we are doing as we are taking a limit is that we are finding out what happens to the function as you make x get as close as possible to 1, in other words, making our , the distance from 1, very small.

All that is being said by is that we are starting out at 1 and making it smaller (depending on ) - in other words, that we can't let our distance from get too big, because this will make the much too big, and we are only worrying about what happens near anyway.

Ahh, yeah agree with the neighborhood being (0,2) and I now see what you mean about min{1, epsilon/4}. So it shouldn't be delta = 1, but rather delta <_ 1 and, using the conditions this imposes on what 1 / |x+4| can be, we say delta = min{1, epsilon/4} to satisfy both the conditions we've assumed...yes?

Also, how're you making the typeface pretty? I hate having to write out delta, epsilon, sigma, etc.

Thanks so much for the help...analysis not so easy...
Mariogs