Continuity of a Rational Function at a number help

kashan123999
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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
 
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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.
 
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".
 
HallsofIvy said:
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...
 
kashan123999 said:

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).
 
Ray Vickson said:
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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