MATHEMATICS · Multi-part · AP · Calculus AB · Limits and Continuity · Limits and Continuity
Problem
Consider the function f(x)=x2+1. Show that x→3limf(x)=10 by establishing an explicit relationship between ε and δ.
(a) Factor and simplify ∣f(x)−10∣ to obtain an expression that involves ∣x−3∣ and ∣x+3∣.
(b) Restrict attention to values of x satisfying ∣x−3∣<1. Under this restriction, determine an upper bound for ∣x+3∣ and use it to produce an inequality of the form ∣f(x)−10∣⩽K∣x−3∣, where K is a positive constant you must specify.
(c) For a given ε>0, choose a positive δ that depends on ε and the constant K so that ∣f(x)−10∣<ε whenever ∣x−3∣<δ and ∣x−3∣<1. State the explicit formula for δ.
(d) Justify why the additional condition ∣x−3∣<1 does not limit the generality of the proof, and conclude that the desired limit equals 10. Provide a concise logical explanation rather than further algebra.