Let f(x) = x^2−9/|x−3|. Show that limx→3f(x) does not exist. (Do not use a table ofvalues or any graphs.)

Answer 1

Since x² - 9 = (x+3) (x -3),

It can be seen that with a change of variable , the limit in question is the same as the limit as y approaches 0 of

(y + 6) y/ | y |

This limit will not exist if the limit of y / | y | does not exist.

The one sided limit as y approaches 0^- is -1, while the one sided limit as y approaches 0⁺ = 1

Since the one sided limits do not agree, the limit as y approaches 0 does not exist by definition,

Answer 2

We consider the limit as x approaches 3 form the left, x→3^- (2.99, 2.999, 2.9999, etc..)

In this case the denominator |x-3| is negative and hence we need to evaluate the limit as x→3^- of f(x) where f(x)=x² -9 / (-x+3)=-(x-3)(x+3)/(x-3)=-(x+3) . The limit in this case is equal to -6.

On the other hand we can evaluate the limit as x approaches 3 from the right , x→3⁺ (3.01, 3.001, 3.0001, etc...). In this case the denominator |x-3| is positive and we can remove the absolute value without changing the sign. Thus, we evaluate the limit as x→3⁺ of f(x) where f(x)=x²-9 / (x-3)=(x-3)(x+3)/(x-3)=(x+3) . The limit in this case is equal to 6.

We note that the two limits are not equal, and hence the limit doesn't exist.

Best,

Davide