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

Question

1729 views around the world

You can reuse this answer
Creative Commons License

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#