How do you determine the limit of #1/(x-2)^2# as x approaches 2?

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Answer 1 · 2018-06-12T10:44:05

For #x->2# we have that #(x-2)^2# can be made as small as we want and is always positive, so that:

#1/(x-2)^2#

can be made as large as we want and is always positive.

Then:

#lim_(x->2) 1/(x-2)^2 =+oo#

Formally, for any #M >0# choose #delta_epsilon < sqrt(1/M)#. Then for #x in (2-delta_epsilon,2+delta_epsilon)# we have that:

# abs(x-2) < delta_epsilon < sqrt(1/M)#

and so:

#(x-2)^2 < 1/M#

and:

#1/(x-2)^2 > M#

which proves the limit.