Question

What is the removable and nonremovable discontinuities in the equation `(x-2)/(x^2 + x - 6)`?

Answer 1

The discontinuity you can't remove is at `x=-3`, because it is the zero of the denominator only.

But discontinuity at 2 can be removed, because at this point the value is not defined (both numerator and denominator are 0), but the limit `lim_(x->2)f(x)` exists and is equal to `1/5`, so if you define a function:

`g(x)={ (((x-2)/(x^2+x-6)) if x!=2), (1/5 if x=2) :}`

it is still discontinuous at -3, but it is continuous at 2