Question

Which function has a point of discontinuity at x=3? A) x-3/2x^2 -2x -12 B) x+3/x^2 -6x +9. Please Explain why you chose the answer.

Answer 1

B has discontinuity at `x=3`

Explanation:

(A) In the function `(x-3)/(2x^2-2x-12)`, both numerator and denominator are equal to zero when `x=3`.

Then what happens when `x->3`. For this let us find

`lim_(x->3)(x-3)/(2x^2-2x-12)`

= `lim_(x->3)(x-3)/(2x^2-6x+4x-12)`

= `lim_(x->3)(x-3)/(2x(x-3)+4(x-3)`

= `lim_(x->3)(x-3)/((2x+4)(x-3))`

= `lim_(x->3)1/(x+4)`

= `1/7`

Hence though `(x-3)/(2x^2-2x-12)` is not defined at `x=3`, we can still have value of `(x-3)/(2x^2-2x-12)` at `x=3` andhence, it is continuous.

(B) The function `(x+3)/(x^2-6x+9)=(x+3)/(x-3)^2`

Hence as `x->3`, though nummerator tends to `6`, denominator is `0` and hence

`lim_(x->3)(x+3)/(x-3)^2=oo`

Hence `(x+3)/(x^2-6x+9)` is not continuous at `x=3`

i.e. (B) has discontinuity