# I've gone full retard stupid continuity question

1. Sep 23, 2010

### charity4thep

is the equation f(x)=(x^2-1)/(x+1) continuous?

i know it can be reduced to f(x)=(x-1) but i remember that in doing so you divide by zero for x=-1 and thus it will be discontinuous at that point...

i dunno i'm really tired tonight

2. Sep 23, 2010

### novop

Never go full retard...

The equation is discontinuous when the denominator is zero.

3. Sep 23, 2010

### charity4thep

thanks man. been a while since i had calc 1 i don't remember the exact rule of this situation. doesn't help that my high school calc teacher taught me a complete 180 from what my university professor did...

4. Sep 24, 2010

### losiu99

Definition of continuity requires a function to be defined in point in which it is continuous.

Last edited: Sep 24, 2010
5. Sep 24, 2010

### HallsofIvy

by the way, the problem is not to determine if the "equation" is continuous- it is to determine if the function defined by that equation is continuous. "continuity" is defined for functions, not equations.

The definition of "f(x) is continuous at x= a" has three parts:
1) That f(a) exist.
2) That $\displaytype \lim_{x\to a} f(x)$ exist.
3) That $\displaytype \lim_{x\to a} f(x)= f(a)$.

As losiu99 says, $(x^2- 1)/(x- 1)$ is not defined at x= 1 and so is not continuous there. $(x^2- 1)/(x-1)= x+ 1$ for x not equal to 1 and is not defined at x= 1. Its graph is NOT the straight line y= x+ 1, it is the straight line y= x+ 1 with a hole at (1, 2).-