# Continuity of a Rational Function at a number help

1. Aug 16, 2013

### kashan123999

1. The problem statement, all variables and given/known data

Find continuity of function f(x)= (x^2-1)/(x-1) at x = 1

2. Relevant equations

limit f(x) as x-> = L

3. The attempt at a solution

i KNOW it can be easily solved by stating that at x = 1 function becomes infinity,so discontinous it is actually...But as we do in finding domain and range of rational functions,we make an altered functions by factorization...in this case by factorization f(x) = x+1 which when x = 1 give limit = 2....I couldn't understand the notion here why we don't factorize here?? please tell me thoroughly in layman's terms

Last edited: Aug 16, 2013
2. Aug 16, 2013

### CAF123

You have the limit right. Only for $x \neq 1,$ does $$\frac{x^2 - 1}{x-1} = x + 1$$

Since you are not interested in what happens to the function at x=1 (even though you know it is discontinuous and the value of f(1) does not exist) you may factorise to find the limit as you have done.

3. Aug 16, 2013

### HallsofIvy

The answer to your question, whether $(x^2- 1)/(x- 1)$ is continuous at x= 1 or not is, clearly, that is it NOT continuous there. It is not even defined at x= 1! (However, the function does NOT "become infinity" at x= 1 nor does it "go to infinity" as x approaches 1.)

IF the question were, instead, whether the function defined as "$f(x)= (x^2- 1)/(x- 1)$ if x is NOT equal to 1, f(1)= 2" is continuous at x= 1, then the answer is "yes".

4. Aug 16, 2013

### kashan123999

but if we factorize it the limit becomes 2...

5. Aug 16, 2013

### Ray Vickson

The function that you wrote, which is
$$f(x) = x^2 - \frac{1}{x} -1$$
is obviously finite and continuous at x=1.

I interpreted the question exactly as written and when read/parsed by standard rules. If you had meant
$$f(x) = \frac{x^2-1}{x-1}$$
you would have know to use brackets, like this: f(x) = (x^2-1)/(x-1).

6. Aug 16, 2013

### kashan123999

pardon my brain-fart sire