Calculus Limits & Continuity Help

Explain why the function is discontinuous at a=1.
Sketch the graph of function.

f(x) = (x^2 - x) / (x^2 - 1) if x does not equal to 1
f(x) = 1 if x = 1

The fact that the denominator of (x^2 - x) / (x^2 - 1) is zero when x = 1 alone does not imply that the f(x) is discontinuous at x = 1, since f(1) is actually defined (because we are told that f(x) = 1 if x = 1).

To show that f(x) is discontinuous at x = 1, we will need to show that
lim x --> 1 of f(x) is unequal to f(1).

lim x --> 1 of f(x)
= lim x --> 1 of (x^2 - x) / (x^2 - 1)
= lim x --> 1 of x(x - 1) / [(x - 1)(x + 1)]
= lim x --> 1 of x / (x + 1)
= 1/2, which is unequal to f(1) = 1.

So f(x) is discontinuous at x = 1.

Lord bless you today!

The function is discontinuous at a =1 because if you put 1 into the denominator function it would equal to zero. This would make the overall function unoperable at that point, and if you were to graph the function on a calculator, you would see a sharp drop and sudden rise where it 'jumped over' the x value of 1.

it's discontinuous at x=1 because it forms an asymptote there. asymptotes are not continuous, they just go to infinity at what ever value. and anytime you have something like x/(1-x), you'll get an asymptote at whatever point the denominator equals zero

lim { x---> 1 } f(x) = 1 / 2 ╪ 1 = f(1)...thus a jump but removable discontinuity at x = 1