In the limit #lim 5/sqrt(2-x)=oo# as #x->2^-#, how do you find #delta>0# such that whenever #2-delta<x<2#, #5/sqrt(2-x)>100#?

Question

1097 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-08T15:45:10

Solve the inequality for #x#, then find #delta#.

Since #sqrt(2-x)# is positive, we can mu;tiply without changing the inequality, so we want

#5/100 > sqrt(2-x)#

or

#sqrt(2-x) < 0.05#

Both are positive and if #a > b# and both are positive, then #a^2 > b^2#. So we want

#2-x < 0.0025# and

#2-0.0025 < x#

#delta = 0.0025#