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

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Answer 1 · 2017-10-22T00:15:53

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#