Proving 0 is Limit of x^2-4 as x→2 via Epsilon-Delta Def.

[JG89]

Homework Statement

Prove using the epsilon-delta definition of a limit that 0 is the limit as x approaches 2 of x^2 - 4

Homework Equations

The Attempt at a Solution

I've never actually done a limit proof like this before, so I just want to make sure that it's correct.

$$|x^2 - 4| = |x-2||x+2| < \delta |x+2|$$.

We can restrict the size of our delta-interval small enough so that for a fixed quantity c, $$\delta < c$$ and 0 <= 2 - c. Since $$\delta < c$$ then $$2 - c < 2 - \delta$$ and $$2 + \delta < 2 + c$$, and so $$2 - c < x < 2 + c \Rightarrow |x + 2| < |4 + c|$$, and so $$\delta |x+2| < \delta |4 + c| = \epsilon$$ and taking $$\delta = \frac{\epsilon}{4+c}$$ shows that we can always find a suitable delta such that for each epsilon > 0, |x^2 - 4| < epsilon whenever |x-2| < delta

How's this look?

 

[snipez90]

If you are going to restrict delta, you might as well pick a specific value. For instance, we can suppose delta is less than or equal to 1 so that |x-2| < 1 implying 1 < x < 3 so 3 < x + 2 < 5 and |x+2| < 5. Then we have d|x+2| < 5d. Letting d = min{1, e/5} works.

Or, you can also note that |x+2| = |x -2 +4| =< |x-2| + 4 by the triangle inequality. Then you need d(d+4) = e, so you can use the quadratic formula to find delta as a function of epsilon.

 

[moomookow]

true, but you can assume c=1 (that is |x-2|<1 ) at the first place to simplify the proof. (so that |x-2||x+2|<5|x-2|<e, if |x-2|<e/5 )

 

[JG89]

snipez, I like your second method of finding a suitable delta. Thanks!

 

[HallsofIvy]

true, but you can assume c=1 (that is |x-2|<1 ) at the first place to simplify the proof. (so that |x-2||x+2|<5|x-2|<e, if |x-2|<e/5 )

In order to assume |x-2|< 1 you must say $\delta$= min(1, $\epsilon$/5)- the smaller of the two numbers.