#lim_{x->3} 1/(x-3) != L# means that there is an #epsilon > 0#, such that for any #delta > 0#, there is an #0 < abs(x - 3) < delta# so that #abs(1/(x-3) - L) >= epsilon#.
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To use the above to show that #lim_{x->3} 1/(x-3)# does not exist, it is sufficient to prove that #lim_{x->3} 1/(x-3) != L# for any real number #L#.
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Let #epsilon = 1# (arbitrarily).
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First, consider #L >= 0#.
Notice that when #x = 3 + 1/(1 + L) > 3#
#abs(1/(x-3) - L) = 1 = epsilon#
If #x = 3 + 1/(1 + L)# is not in the interval #0 < abs(x - 3) < delta#, then #delta <= 1/(1 + L)#.
This also means that
#1/((3 + delta/2)-3) - L > 1/((3 + delta)-3) - L#
#>= 1/((3 + 1/(1 + L))-3) - L#
#= 1 = epsilon#
and #x = 3 + delta/2# is guaranteed to be in the interval #0 < abs(x - 3) < delta#.
Thus, let #x = min( 3 + delta/2, 3 + 1/(1 + L))#.
This ensures that #0 < abs(x - 3) < delta# and #abs(1/(x-3) - L) >= epsilon#, and therefore #lim_{x->3} 1/(x-3) != L#.
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Next, consider #L < 0#.
Similarly, notice that when #x = 3 - 1/(1 - L) < 3#
#abs(1/(x-3) - L) = abs(-1) = 1 = epsilon#
If #x = 3 - 1/(1 - L)# is not in the interval #0 < abs(x - 3) < delta#, then #delta <= 1/(1 - L)#.
This also means that
#1/((3 - delta/2)-3) - L < 1/((3 - delta)-3) - L#
#<= 1/((3 + 1/(1 - L))-3) - L#
#= -1#
or
#abs(1/((3 - delta/2)-3) - L) > 1 = epsilon#
and #x = 3 - delta/2# is guaranteed to be in the interval #0 < abs(x - 3) < delta#.
Thus, let #x = max( 3 - delta/2, 3 - 1/(1 - L))#.
This ensures that #0 < abs(x - 3) < delta# and #abs(1/(x-3) - L) >= epsilon#, and therefore #lim_{x->3} 1/(x-3) != L#.
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Hence, #lim_{x->3} 1/(x-3) != L# for all #L# and therefore, #lim_{x->3} 1/(x-3)# does not exist.
