In the limit #lim sqrt(x^2-4)=0# as #x->2^+#, how do you find #delta>0# such that whenever #2<x<2+delta#, #sqrt(x^2-4)<0.01#?
1 Answer
Apr 7, 2018
Explanation:
Since the convergence to the limit is monotonic, we can solve:
#sqrt(x^2-4) = 0.01#
to find the upper limit for
Squaring both sides, this becomes:
#x^2-4 = 0.0001#
Add
#x^2 = 4.0001#
Then taking the positive square root of both sides, we get:
#x = sqrt(4.0001) ~~ 2.00002499984375195309#
So we could choose