How do you find the lim #(x^2-x+12) / (x+3)# as x approaches -3?

1 Answer
Aug 4, 2015

#lim_(xrarr-3)(x^2-x+12)/(x+3)# does not exist.

Explanation:

As #xrarr-3#,

we get the numerator: #x^2-x+12 rarr 9+3+12 = 24#

while the denominator: #x+3 rarr 0#

There is no number that the ratio is getting closer to.

In fact If #x# is close to #-3# but a little left of #-3#, then the denominator is a negative fraction very close to #0#, so the whole thing is a very big negative number.

We write: #lim_(xrarr-3^-)(x^2-x+12)/(x+3) = -oo# to indicate that the limit as #x# approaches #-3# from the left does not exist because as #x# approaches #-3# from the left, the ratio decreases without bound.
(Yes, that is an awful lot to write. That's why it's nice to have a good notation for it.)

A similar analysis on the right of #-3# will show us that limit as #x# approaches #-3# from the right does not exist because as #x# approaches #-3# from the right, the ratio increases without bound.

We write: #lim_(xrarr-3^+)(x^2-x+12)/(x+3) = oo#

In my experience in the US, for the two sided limit, we write:

#lim_(xrarr-3)(x^2-x+12)/(x+3)# does not exist.

or #lim_(xrarr-3)abs((x^2-x+12)/(x+3))=oo#

I have been told that the convention in other places is to write
#lim_(xrarr-3)(x^2-x+12)/(x+3)=oo# in this case.