How do you prove that the limit of #1/(x-3)# does not exist as x approaches 3 using the epsilon delta proof?

Question

Charles Anctil · Accepted Answer

#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#. # # # #

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#. # # # #

Let #epsilon = 1# (arbitrarily). # # # #

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#. # # # #

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#. # # # #

Hence, #lim_{x->3} 1/(x-3) != L# for all #L# and therefore, #lim_{x->3} 1/(x-3)# does not exist.

Evelyn Agard · Answer

undefined To prove that the limit of 1/(x-3) does not exist as x approaches 3 using the epsilon-delta proof, we need to show that for any given epsilon value, there does not exist a corresponding delta value that satisfies the definition of a limit.

Let's assume that the limit L exists. According to the epsilon-delta definition, for any epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 3| < delta, then |1/(x-3) - L| < epsilon.

To disprove the existence of the limit, we need to find an epsilon value for which there is no corresponding delta value. Let's consider the case when epsilon is equal to 1.

Now, we need to show that for any delta value chosen, there exists an x value such that 0 < |x - 3| < delta, but |1/(x-3) - L| ≥ 1.

Let's assume L is a real number. If we choose x = 3 + delta/2, then |x - 3| = delta/2 < delta. However, |1/(x-3) - L| = |1/(delta/2) - L| = |2/delta - L| ≥ 1, which contradicts the definition of the limit.

Therefore, we have shown that for epsilon = 1, there does not exist a corresponding delta value that satisfies the definition of a limit. Hence, the limit of 1/(x-3) does not exist as x approaches 3 using the epsilon-delta proof.