Interactive Real Analysis - part of MathCS.org

• Real Analysis
• 1. Sets and Relations
• 2. Infinity and Induction
• 3. Sequences of Numbers
• 4. Series of Numbers
• 5. Topology
• 6. Limits, Continuity, and Differentiation
• 6.1. Limits
• 6.2. Continuous Functions
• 6.3. Discontinuous Functions
• 6.4. Topology and Continuity
• 6.5. Differentiable Functions
• 6.6. A Function Primer
• 7. The Integral
• 8. Sequences of Functions
• 9. Historical Tidbits
• Java Tools

Proposition 6.1.6: Equivalence of Definitions of Limits

Why these ads ...

If f is any function with domain D in R, and c closure(D) then the following are equivalent:

1. For any sequence {x_n} in D that converges to c the sequence {f(x_n)} converges to L
2. given any > 0 there exists a > 0 such if x closure(D) and | x - c | < then | f(x) - L | <

Context

Proof:

Suppose the first condition is true, but the second condition fails. Then there exists an > 0 such that there is no > 0 with the property that if | x - c | < then | f(x) - L | < . Therefore, if we let = 1 / n, then for each n we can produce a number x_n with

• | x_n- c | < but | f(x_n) - L |

But then the sequence { x_n} converges to c, but the sequence f(x_n) does not converge to L. That is contrary to the first condition being true, and hence we have proved by contradiction that the first condition implies the second.

Suppose the second condition is true. Let c be some number in closure(D) and pick any > 0. There exists a number > 0 such that

• whenever | x - c | < then | f(x) - L | <

Take any sequence { x_n} in D that converges to c. Then there is an integer N such that

• | x_n- c | < for n > N

But then, by assumption of the second condition,

• | f(x_n) - L | < for n > N

But that is the definition of the sequence { f(x_n) } converging to L, as required.