# Confirm Precise Definition of a Limit solution

by kylera
Tags: confirm, definition, limit, precise, solution
 P: 40 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 $$\delta$$ Since 0 < |x + 2|<$$\delta$$, |f(x) - 3| < $$\epsilon$$ |f(x) - 3| = |x^2 - 1 - 3| = |x^2 - 4| = |x + 2||x - 2|< $$\epsilon$$ Let C = |x - 2|, which leads to C|x + 2| < $$\epsilon$$ |x + 2| = $$\delta$$ < $$\frac{\epsilon}{C}$$ Applying the Precise Definition of a Limit, For a $$\delta$$ value $$\frac{\epsilon}{C}$$ greater than zero, there exists $$\epsilon$$ greater than zero such that if 0 < |x + 2|<$$\delta$$, then |f(x) - 3| < $$\epsilon$$. |f(x) - 3| < $$\epsilon$$ |f(x) - 3| = |x^2 - 1 - 3| = |x^2 - 4| = |x + 2||x - 2|< $$\epsilon$$ Re-apply C and $$\delta$$ to get C x $$\frac{\epsilon}{C}$$ < $$\epsilon$$ Hence, by the Precise Definition of a limit, said limit does exist. Much thanks in advance. Comments and criticisms are always welcome.
 P: 152 It looks like you're doing this backwards. Your proof should start like "Given $\epsilon > 0$ there exists $\delta > 0$ such that if $0<|x+2|<\delta$, then $|f(x)-3|<\epsilon$..." Now show that this is true using the $\delta$ that you found.
Math
Emeritus
 Quote by kylera 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 $$\delta$$ Since 0 < |x + 2|<$$\delta$$, |f(x) - 3| < $$\epsilon$$ |f(x) - 3| = |x^2 - 1 - 3| = |x^2 - 4| = |x + 2||x - 2|< $$\epsilon$$ Let C = |x - 2|, which leads to C|x + 2| < $$\epsilon$$
 |x + 2| = $$\delta$$ < $$\frac{\epsilon}{C}$$ Applying the Precise Definition of a Limit, For a $$\delta$$ value $$\frac{\epsilon}{C}$$ greater than zero, there exists $$\epsilon$$ greater than zero such that if 0 < |x + 2|<$$\delta$$, then |f(x) - 3| < $$\epsilon$$. |f(x) - 3| < $$\epsilon$$ |f(x) - 3| = |x^2 - 1 - 3| = |x^2 - 4| = |x + 2||x - 2|< $$\epsilon$$ Re-apply C and $$\delta$$ to get C x $$\frac{\epsilon}{C}$$ < $$\epsilon$$ Hence, by the Precise Definition of a limit, said limit does exist. Much thanks in advance. Comments and criticisms are always welcome.