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#?

Since the convergence to the limit is monotonic, we can solve:

Squaring both sides, this becomes:

Then taking the positive square root of both sides, we get:

To find the value of delta, we can start by solving the inequality sqrt(x^2-4) < 0.01.

First, we square both sides of the inequality to eliminate the square root: x^2 - 4 < 0.01^2.

Simplifying this inequality, we get x^2 - 4 < 0.0001.

Next, we add 4 to both sides of the inequality: x^2 < 4.0001.

Taking the square root of both sides, we have |x| < sqrt(4.0001).

Since we are interested in the interval (2, 2 + delta), we can ignore the negative values of x.

Therefore, we have 2 < x < 2 + delta.

Substituting this into the inequality, we get 2 < 2 + delta < sqrt(4.0001).

Simplifying further, we have 0 < delta < sqrt(4.0001) - 2.

Calculating the value of sqrt(4.0001) - 2, we find that it is approximately 0.001.

Therefore, we can choose delta = 0.001 as a suitable value such that whenever 2 < x < 2 + delta, sqrt(x^2-4) < 0.01.