Question

How do you find the limit of `(abs(x+2)-3)/(x-7 )` as x approaches 7?

Answer 1

`lim_(x to 7^+) (abs(x+2)-3)/(x-7 ) = oo`

`lim_(x to 7^-) (abs(x+2)-3)/(x-7 ) = -oo`

Explanation:

`lim_(x to 7) (abs(x+2)-3)/(x-7 )`

`= (lim_(x to 7) (abs(x+2)-3))/(lim_(x to 7)x-7 )`

and because the numerator is continuous

`= (6)/(lim_(x to 7)x-7 )`

Note that this gives rise to a singularity and a 2 sided limit.

To test this in your head, mentally plug in, say, `x = 6.9` and `x = 7.1` So the right-sided limit is positive and the left-sided limit is negative

So we say that:

`lim_(x to 7^+) (abs(x+2)-3)/(x-7 ) = oo`

`lim_(x to 7^-) (abs(x+2)-3)/(x-7 ) = -oo`

Answer 2

The limit does not exist.

Explanation:

Simple substitution of `x = 7` gives a numerator of 6 and a denominator of 0. There would be division of zero.

However, if `x` approaches 7 from the left, then the expression tends to negative infinity. We write

`lim_{x -> 7^-} frac{abs(x+2)-3}{x-7} = -oo`

Similarly, if `x` approaches 7 from the right, then the expression tends to infinity.

`lim_{x -> 7^+} frac{abs(x+2)-3}{x-7} = oo`