How do you prove the limit of sqrt (x) = 3 as x approaches 9 using the epsilon-delta definition?

1 Answer
Andrea S.
Feb 23, 2018

Given any #epsilon > 0# choose #delta_epsilon# such that:

#delta_epsilon < 3epsilon#

Now consider #x in (9-delta_epsilon, 9+delta_epsilon)# and evaluate the difference:

#abs (sqrtx-3)#

Multiply and divide by #sqrtx + 3 > 0#

#abs (sqrtx-3) = abs ((sqrtx-3)(sqrtx+3))/(sqrtx+3)#

and as #(a+b)(a-b) = a^2-b^2# we have:

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

The highest possible value of the numerator for #x in (9-delta_epsilon, 9+delta_epsilon)# is #delta_epsilon#, while the lowest possible value for the denominator is #sqrt(9-delta_epsilon) + 3 > 3# and then:

#abs (sqrtx-3) < delta_epsilon/3 #

but then as #delta_epsilon < 3epsilon#

#abs (sqrtx-3) < epsilon #

which proves the point.

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