Thread: ε-δ Limit with an x² term?

1. ε-δ Limit with an x² term?

Hello , I have been having trouble with understanding what to do with those ε-δ Limits that involve an extra "x" factor in the result, I'll just write out the problem and perhaps you could correct or instruct me on the logic behind it.

$\displaystyle \lim_{x \to 2}(x^2 + x - 2) = 4$

$\displaystyle 0 < |x - 2| < \delta \Rightarrow |(x^2 + x - 2) - 4| < \epsilon$

$\displaystyle 0 < |x - 2| < \delta \Rightarrow |x^2 + x - 6| < \epsilon$

$\displaystyle 0 < |x - 2| < \delta \Rightarrow |(x + 3)(x - 2)| < \epsilon$

$\displaystyle 0 < |x - 2| < \delta \Rightarrow |x + 3||x - 2| < \epsilon$

Here, I'm kind of confused. I've seen the best explanation so far here: Created with Camtasia Studio 5 & I'd like to follow the method to get everything in working order.

1: Because of the |x - 2|< δ, (i.e. the x-->2 in the original limit), we know that x is very close to 2, just a bit bigger than it. Because this is so the |x + 3| term is very close to 5, just a bit above it. In fact, it will definitely be less than 6.

We can say with confidence that;

|x + 3||x - 2| < 6|x - 2|

2: Here I'm a bit unsure, can we say that;

6|x - 2| < ε

|x - 2|< ε/6 ? I think this works because although the gap from. say, 5.0000001 to being less than 6 (from the inequality) is big, when we divide ε by 6 we are creating more space.

3: Is there a way to make it look pretty? Is the answer that as long as δ = ε/6 we can guarantee the limit?

2. I recommend you read this for detailed explanation and working.

http://www.mathhelpforum.com/math-he...ta-proofs.html

You'll find example 3 to be especially helpful.

3. I already had that pdf on my laptop & I never understood it until I followed the video I linked to but I may understand it now.

If I take my example;

$\displaystyle |x + 3||x - 2| < \epsilon$

$\displaystyle |x - 2| < \frac{ \epsilon}{ |x + 3|}$

What I can do is look back at |x - 2| & if I want to bound it within 1 unit it would be equivalent to bound the |x + 3| factor in the denoinator within a distance of 1 as well;

$\displaystyle |x - 2| < 1$

$\displaystyle -1 < x - 2 < 1$

Then, because the limit is going near 2 we'll take the |x + 3| and realise it's near 5, but because of the constriction in |x - 2| of being within 1 unit we'll do the same for the |x + 3|

$\displaystyle -1 < x - 2 < 1 \Rightarrow 4 < x + 3 < 6$

Then, if we take 6 as the min. i.e. δ = (1,ε/6)

$\displaystyle |x + 3||x - 2| < \epsilon$

$\displaystyle 6|x - 2| < \epsilon$

$\displaystyle |x - 2| < \frac{ \epsilon}{6}$

$\displaystyle |x - 2| < \frac{ \epsilon}{6}$

So;

$\displaystyle |x + 3||x - 2|$

$\displaystyle 6|x - 2|$

$\displaystyle 6 ( \frac{ \epsilon}{6})$

$\displaystyle \epsilon$

I haven't made any careless mistakes or assumptions have I?