How do you find the x values at which #f(x)=x/(x^2-x)# is not continuous, which of the discontinuities are removable?

Question

Christopher Allers · Accepted Answer

A rational function is continuous on its domain and only on its domain.

The domain of #f# is all real #x# except solutions to #x^2-x=0#. Since the solutions are #0# and #1#, the function #f# is not continuous at #0# and at #1#. A discontinuity of function #f# at #a# is removable if and only if #lim_(xrarra)f(x)# exists. Checking the discontinuity at #0#: #lim_(xrarr0) x/(x^2-x) = lim_(xrarr0)1/(x-1) = -1#. This discontinuity is removable. Checking the discontinuity at #1#: #lim_(xrarr0) x/(x^2-x) = lim_(xrarr0)1/(x-1) # does not exist. This discontinuity is not removable. Additional facts #lim_(xrarr1^-)1/(x-1) = -oo# and #lim_(xrarr0^+)1/(x-1) = oo# A discontinuity like this is sometimes called an infinite discontinuity. Infinite discontinuities are not removable.

Ezra Adams · Answer

undefined To find the x values at which f(x) = x/(x^2-x) is not continuous, we need to identify the points where the function is undefined or where the limit does not exist.

First, we observe that the function is undefined when the denominator, x^2-x, equals zero. Solving x^2-x=0, we find that x=0 and x=1. Therefore, these are potential points of discontinuity.

Next, we check the limits at these points to determine if they exist. Taking the limit as x approaches 0, we have lim(x→0) x/(x^2-x). Simplifying, we get lim(x→0) 1/(x-1). This limit does not exist, so the discontinuity at x=0 is not removable.

Similarly, taking the limit as x approaches 1, we have lim(x→1) x/(x^2-x). Simplifying, we get lim(x→1) 1/(x-1). This limit also does not exist, so the discontinuity at x=1 is not removable.

In summary, the function f(x) = x/(x^2-x) is not continuous at x=0 and x=1, and both of these discontinuities are not removable.