## Princeton Review 3rd Ed - Error regarding "Continuity"

Forum for the GRE subject test in mathematics.
Legendre
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### Princeton Review 3rd Ed - Error regarding "Continuity"

Page 38

f(x) = (x^2 - 1) / (x -1)

He says f is discontinuous at x = 1 because f is not defined at this point.

From what I know :

We only talk about the continuity of a function over its domain. f do not have x = 1 in its domain because a function is a rule that assigns each element in its domain to exactly one element in its co-domain.

So f is in fact continuous over the largest subset of R on which it is defined. We cannot discuss the continuity of f on domain R because R cannot be f's domain.

Am I right?

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

### Re: Princeton Review 3rd Ed - Error regarding "Continuity"

Yes, you are right Legendre
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### Re: Princeton Review 3rd Ed - Error regarding "Continuity"

Nameless wrote:Yes, you are right Thanks!

zombie
Posts: 27
Joined: Thu Nov 20, 2008 2:30 am

### Re: Princeton Review 3rd Ed - Error regarding "Continuity"

I just checked that, and it's a really bad example, because you are quite right, when written in that form, the function is not discontinuous at x = 1, its domain is all of R\{1}, yet f is not asymptotic at 1, it is indeterminate.

This function is also equivalent to the line f(x) = x + 1 which is bijective and continuous and homeomorphic (i.e. has all the nice sorts of topological properties that a straight line has in the Euclidean plane)

In the context of Calculus and Real Analysis though, the left/right limits exist and are equal at 2. Recall, that the limit of a function can exist at a point which is in a deleted neighborhood of an interval, such as the point {1} in R.

Yet, a function is continuous at a point iff the limit of the function at the point is equal to the value of the function at that point, f(1), which is clearly undefined.

That example is quite confusing because with a little algebra, it is continuous. It's like saying "The square root of x is discontinuous in the second quadrant" which makes absolutely no sense unless you mean on the complex plane.

Given that we know far beyond the basics of first semester Calculus, what he means in that particular context is that if f is a function with domain R and codomain R

f: R -> R

where f is a relation (x,y), and at 1, the relation could only be written as (1, und), the necessary and sufficient condition for continuity does not hold true... it just could not. So, there's only two possible bins to put the behavior of the function f at the point 1 and that is either continuous or discontinuous. In the context, DeLuc just haphazardly throws it into the discontinuous bin without any regard for the indeterminate behavior of the function at the point.

Normally, of course, we would not write that undefined is in the range of the function on R and we would write that the range is the image of some set A = R\{1} under f.

I would say that it's always a slight error (or unclear statement) when asymptotic or indeterminate behavior is considered discontinuous, because it's more accurate and descriptive to say that a function is continuous on its domain and then state what that domain is as a subset of the reals. However, it should also be assumed and recognized that many authors just throw undefined points and asymptotic behavior into the "discontinuous bin" justified implicitly on the "law of the excluded middle".

I know, I'm rambling, yet this got me thinking because it's more of a special case than say
\frac{1}{x} at 0 where \lim_{x \to\ 0}\frac{1}{x} does not exist. In such a case, the question of continuity of the function at x = 0 is absurd.

(Also, does Tex work on this message board??)