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

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Answer 1 · 2016-10-09T08:53:56

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

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

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

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

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 · 2016-10-09T08:56:42

The limit does not exist.

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.

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

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