How do you evaluate the limit #1/x^2# as x approaches #0#?

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Answer 1 · 2017-02-06T01:08:34

#lim_(x->0) 1/x^2 = +oo#

This is quite evident, since, for #x->0#, #x^2# is positive and indefinitely small, so its reciprocal is positive and indefinitely large.

However, it can be proved easily in the #delta-epsilon# form:

GIven any #M > 0# we can choose #delta_M = 1/sqrt(M)#.

So, for #x in (-delta_M, delta_M)# we have:

#abs(x) < 1/sqrt(M)#

#x^2 < 1/M#

#1/x^2 > M#

which proves the point.