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

1 Answer
Jun 12, 2018

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.