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?