How do you determine the limit of #(2)/(x-3)# as x approaches #3^-#??

1 Answer
May 25, 2016

The expression #2/(x-3)# tends to negative infinity as #x# approaches #3^-# (that is, as #x->3# while being always less than #3#)
See comments below.

Explanation:

Recall the definition of a limit:

We say that the limit of function #f(x)# (defined for all #x# in some neighborhood, however small, of value #x=a#, but not necessarily at point #x=a# itself) is #A# as #x->a#
(written as #lim_(x->a) f(x) = A# or #f(x)_(x->a)->A#)
if and only if for any #epsilon > 0# (measuring the closeness of #f(x)# to value #A#)
there is such #delta > 0# (measure of the closeness of #x# to value #a#) that
for any #x# satisfying the inequality #|x-a| < delta#
the following is true: #|f(x)-A| < epsilon#

In other words, for any degree of closeness #epsilon# of #f(x)# to #A# we can find a distance #delta# of #x# to #a# such as the function #f(x)# will be closer to #A# than #epsilon# if #x# is closer to #a# than #delta#.

With infinity the situation needs another definition, because infinity is not a number, like #A# in the above definition.
We should supplement our definition for two cases of infinity - positive and negative.

We say that the limit of function #f(x)# (defined for all #x# in some neighborhood, however small, of value #x=a#, but not necessarily at point #x=a# itself) is negative infinity
(written as #lim_(x->a) f(x) = -oo# or #f(x)_(x->a)->-oo#)
if and only if for any real #A<0# (however large by absolute value)
there is such #delta > 0# (measure of the closeness of #x# to value #a#) that
for any #x# such as #|x-a| < delta#
the following is true: #f(x) < A#

Similarly is defined the limit of positive infinity.

In our case, if #x->3^-# (that is, #x->3# while #x<3#), the fraction #2/(x-3)# is always negative.

Choose any #A<0# and resolve the inequality
#2/(x-3) < A#
the solution is:
#x > 3+2/A#
The larger by absolute value #A# is (while being negative) - the closer #x# should be to #3# for our fraction to be less than #A#.
So, we always find that measure of closeness of #x# to #3# (while being smaller than #3#):
#delta = 2/|A|#
When #x# is closer than this #delta# to #3#, the fraction will fall lower than #A#, no matter what negative #A# we have chosen.