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

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Answer 1 · 2017-01-20T00:18:21

#lim_(xto2)1/(x-2)^2=oo#

Consider the fact that #lim_(xto0^+)1/x=oo#

also notice that

#x=2 => (x-2)^2=(2-2)^2=0^2=0#

So as #xto2# we see that #(x-2)^2 to 0#

so let #u=(x-2)^2#

then

#lim_(uto0^+)1/u=lim_(xto2)1/(x-2)^2#

and since #lim_(xto0^+)1/x=oo#

#lim_(uto0^+)1/u=oo#

and since #u=(x-2)^2# and #uto0# as #xto2#

#lim_(xto2)1/(x-2)^2=oo#