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. 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.
It looks like you're doing this backwards. Your proof should start like "Given [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that if [itex]0<|x+2|<\delta[/itex], then [itex]|f(x)-3|<\epsilon[/itex]..." Now show that this is true using the $\delta$ that you found.
You can't do that. |x-2| is a variable, not a constant what you need is a BOUND on |x-2|. If C is close to -2, say between -1 and -3, how large or how small can |x-2| be?