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

Question

Ezra Adams · Accepted Answer

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.

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.

Emilia Aguilar · Answer

undefined To determine the limit of (2)/(x-3) as x approaches 3^-, we substitute the value of x into the expression. Plugging in x = 3^- gives us (2)/(3^- - 3). Simplifying further, we get (2)/(-∞), which equals 0. Therefore, the limit of (2)/(x-3) as x approaches 3^- is 0.