At first, I thought someone was sending me their calculus homework, but this time of the semester, it’s most likely a question that was asked on a final. Anyway, I don’t mind posting solutions on topics that are of general interest to high-school and college students … whenever I have the time 🙂

Question: What is the limit as x→2 of the function x^{2} + 3 , using only the definition of a limit?

So-called epsilon-delta proofs can be cumbersome for polynomials of degree greater than one because of the apparent lack of a single delta value. To review, limit as x→c of f(x) = f(c) means that for any ε > 0, there exists δ > 0 such that |x – c| < δ ==> |f(c) – f(x)| < ε . Intuitively, the definition means we can make f(x) arbitrarily close to f(c) by making x ‘sufficiently’ close to c.

For the problem at hand, it appears that the limit as x→2 is 7. To prove this assertion using the definition, then given any ε > 0 , we must show there exists a δ > 0 (remember that the delta value is a function of epsilon) such that |f(x) – 7| < ε. Now,

|f(x) – 7| < ε ==> |x^{2} – 4| = | (x+2)(x-2) | = |x+2||x-2| < ε . Notice that lim as x→2 of |x-2| = 0 (this is easily proven from the definition), so the trick is to choose an ‘easy’ bound for one factor and limit the other. Consider two deltas, δ_{1} and δ_{2}. Since x-2 approaches zero as x→2, set δ_{1} = 1. Now,

|x-2| < δ_{1} = 1 ==> -1 < x – 2 < 1 ==> 3 < x + 2 < 5 ==> |x+2| < 5. Take δ_{2} = ε /5 and δ = min(δ_{1}, δ_{2}).

Case 1: Any ε ≥ 5. |x-2| < δ = min(δ_{1}, δ_{2}) ==> |x-2| < 1, so

|f(x)-7| = |x+2||x-2| < 5|x-2| < 5, which is less than ε .

Case 2: Any 0 < ε < 5. |x-2| < δ = min(δ_{1}, δ_{2}) ==> |x-2| < ε/5, so

|f(x)-7| = |x+2||x-2| < 5|x-2| < 5(ε/5) < ε .

Strictly speaking, separate cases were not necessary, but I find that students understand the concept easier the first few times this way.