# Indicators that a limit does/does not exist

1. Sep 19, 2012

### newageanubis

Hey everyone,

When I first started learning calculus, I was taught that the first thing to do when asked to evaluate a limit as x -> a of f(x) is to evaluate f(a). If f(a) is of the form 0/0, then no conclusion can be made about the limit, and the expression needs to be manipulated by factoring, rationalizing, etc. before a conclusion can be made. If f(a) i of the form k/0, where a is a real number and k≠0, then the limit doesn't exist. Finally, if f(a) = k, where k is a real number, then the limit exists and is equal to k. But high school calculus only dealt with functions that behave nicely: polynomials, rational functions, trig functions, exponential/log.

My question is this: does this "method" of discerning the nature of a limit work for all f(x)?

2. Sep 19, 2012

### Number Nine

Consider the following function: Let f(a) = 1 when a is rational, and let f(a) = 0 when a is irrational. What is the limit as a approaches 3?

3. Sep 19, 2012

### newageanubis

I believe that's the Dirichlet function, and I don't think that limit exists...:S

4. Sep 20, 2012

### pwsnafu

Yes on both accounts.

A simpler counterexample would be the step function: f(x) = 0 when x < 0, f(x) = 0.5 when x = 0, and f(x) = 1 when x > 0. The problem with the method lies in
f(a) can always be made to exist, by simply defining f(a)=0, without affecting the limit. In other words, you are making the assumption that f(x) is continuous to begin with, so you test f(a).

Last edited: Sep 20, 2012
5. Sep 20, 2012

### HallsofIvy

Staff Emeritus
In other words, your "indicators" work only for the case where you have a fraction, with one continuous function over another.