In the limit #lim_(x->1^-)1/(1-x^2)=oo#, how do you find #delta>0# such that whenever #1-delta<x<1#, #1/(1-x^2)>100#?

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Answer 1 · 2018-06-10T11:04:34

See below

It must be #1/(1-x^2)>100#

from here #1>100-100x^2# then transposing terms we have

#100x^2>99# then #x^2>99/10# or equivalent #x>+-sqrt(99/100)#.

we choose positive root because #delta>0#

#x>sqrt99/10#

If we take #delta=sqrt99/10# we are sure that:

first.- #delta <1#
second.- if #x in (1-delta,1)# then #1/(1-x^2)>100#