As the subject says, I have "proved" a function, but because this type of problem seems to have no set answer, I'd like some opinions on whether I walked through it properly or not.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

lim(x=> -2) (x^2-1) = 3

2. Relevant equations

Precise Definition of a Limit

3. The attempt at a solution

Part 1. Assume a value for [tex]\delta[/tex]

Since 0 < |x + 2|<[tex]\delta[/tex],

|f(x) - 3| < [tex]\epsilon[/tex]

|f(x) - 3| = |x^2 - 1 - 3|

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

Let C = |x - 2|, which leads to C|x + 2| < [tex]\epsilon[/tex]

|x + 2| = [tex]\delta[/tex] < [tex]\frac{\epsilon}{C}[/tex]

Applying the Precise Definition of a Limit,

For a [tex]\delta[/tex] value [tex]\frac{\epsilon}{C}[/tex] greater than zero, there exists [tex]\epsilon[/tex] greater than zero such that if 0 < |x + 2|<[tex]\delta[/tex], then |f(x) - 3| < [tex]\epsilon[/tex].

|f(x) - 3| < [tex]\epsilon[/tex]

|f(x) - 3| = |x^2 - 1 - 3|

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

Re-apply C and [tex]\delta[/tex] to get

C x [tex]\frac{\epsilon}{C}[/tex] < [tex]\epsilon[/tex]

Hence, by the Precise Definition of a limit, said limit does exist.

Much thanks in advance. Comments and criticisms are always welcome.

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# Homework Help: Confirm Precise Definition of a Limit solution

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