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

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SUMMARY

The function defined by the equation f(x)=(x^2-1)/(x+1) is discontinuous at x=-1 due to the denominator being zero. Although it can be simplified to f(x)=(x-1), the function is not defined at x=1, which also results in discontinuity. The definition of continuity requires that the function be defined at the point in question, and both the limit and the function value must exist and be equal at that point. Therefore, the function is continuous everywhere except at x=-1 and x=1.

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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
 
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Never go full retard...

The equation is discontinuous when the denominator is zero.
 
novop said:
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...
 
Definition of continuity requires a function to be defined in point in which it is continuous.
 
Last edited:
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).-
 

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