In the limit #lim 1/(4-t)=-oo# as #t->4^+#, how do you find #delta>0# such that whenever #4<t<t+delta#, #1/(4-t)<-100#?

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Answer 1 · 2017-11-28T20:59:37

#delta = 1/100#

For any #delta > 0# and #4 < t < t+delta# we have that:

#4-t < 0#

therefore:

#1/(4-t) < -100 => 1 > -100(4-t)#

#1 > -400 +100t#

#t < 401/100#

#t < 4 +1/100#

which shows that we can choose #delta = 1/100#