Is the Equation f(x)=(x^2-1)/(x+1) Continuous? A Stupid Continuity Question

[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 don't know I'm really tired tonight

 

[novop]

Never go full retard...

The equation is discontinuous when the denominator is zero.

 

[charity4thep]

Never go full retard...

The equation is discontinuous when the denominator is zero.

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...

 

[losiu99]

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

 

[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).-