In the limit #lim sqrt(6-3x)=0# as #x->2^-#, how do you find #delta>0# such that whenever #2-delta<x<2#, #sqrt(6-3x)<0.01#?

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Answer 1 · 2017-11-10T16:01:13

Please see below.

Figure out how close #x# must be to #2#. (That is #delta#.)

#sqrt(6-3x) < 1/100#

Square both sides: #6-3x < 1/10^4#

Factor the #3# on the left #3(2-x) < 1/10^4#

So, #2-x < 3/10^4 = 0.0003#

And we need

#2-0.0003 < x#

#delta = 0.0003#