# Formally proving limits with epsilon-delta proofs

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• Apr 30th 2012, 07:14 PM
mariogs379
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
• Apr 30th 2012, 09:04 PM
Prove It
Re: Formally proving limits with epsilon-delta proofs
Quote:

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 \displaystyle \displaystyle \begin{align*} \delta \end{align*}, the distance from 1, very small.

All that is being said by \displaystyle \displaystyle \begin{align*} \delta = \min\left\{1, \frac{\epsilon}{4}\right\} \end{align*} is that we are starting out at 1 and making it smaller (depending on \displaystyle \displaystyle \begin{align*} \epsilon \end{align*}) - in other words, that we can't let our distance from \displaystyle \displaystyle \begin{align*} x = 1 \end{align*} get too big, because this will make the \displaystyle \displaystyle \begin{align*} \epsilon \end{align*} much too big, and we are only worrying about what happens near \displaystyle \displaystyle \begin{align*} x = 1 \end{align*} anyway.
• Apr 30th 2012, 09:39 PM
mariogs379
Re: Formally proving limits with epsilon-delta proofs
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