Continuity of a Rational Function at a number help

[kashan123999]

Homework Statement

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

Homework Equations

limit f(x) as x-> = L

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

 

[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.

 

[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".

 

[kashan123999]

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".

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

limit f(x) as x-> = L

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

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).

 

[kashan123999]

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).

pardon my brain-fart sire