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## 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.

[tex] |x^2 - 4| = |x-2||x+2| < \delta |x+2| [/tex].

We can restrict the size of our delta-interval small enough so that for a fixed quantity c, [tex] \delta < c [/tex] and 0 <= 2 - c. Since [tex] \delta < c [/tex] then [tex] 2 - c < 2 - \delta [/tex] and [tex] 2 + \delta < 2 + c [/tex], and so [tex] 2 - c < x < 2 + c \Rightarrow |x + 2| < |4 + c| [/tex], and so [tex] \delta |x+2| < \delta |4 + c| = \epsilon [/tex] and taking [tex] \delta = \frac{\epsilon}{4+c} [/tex] 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?