How do you prove that the limit of sqrtx = 3 as x approaches 9 using the epsilon delta proof?

1 Answer
Oct 22, 2017

Start from:

abs(x-9) = abs(sqrtx-3)abs(sqrtx+3)

for x > 0, so:

abs(sqrtx-3) = abs(x-9)/(sqrtx+3)

For any number epsilon > 0 choose now delta_epsilon < 3epsilon.
For x in (9-delta_epsilon,9+delta_epsilon) we have;

abs(x-9) < delta_epsilon

and, as sqrtx is positive:

sqrtx +3 > 3

so:

abs(sqrtx-3) = abs(x-9)/(sqrtx+3) < delta_epsilon/3 < (3epsilon)/3 = epsilon

Thus:

x in (9-delta_epsilon,9+delta_epsilon) => abs(sqrtx-3) < epsilon

which proves that:

lim_(x->9) sqrtx = 3