How do you prove that the limit of #(x^2) =9 # as x approaches -3 using the epsilon delta proof?

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Answer 1 · 2017-12-08T22:49:46

See explanation...

Let #epsilon > 0#

Choose #delta in (0, epsilon / 7) nn (0, 1)#

Note that:

#0 < delta^2 < delta#

If #x in (-3-delta, -3+delta)# then:

#abs(x^2 - 9) < abs((-3-delta)^2 - 9) = abs((-3)^2+6delta+delta^2-9) < 7delta < epsilon#

So:

#AA epsilon > 0 EE delta > 0 : AA x in (-3-delta, -3+delta), abs(x^2-9) < epsilon#

So:

#lim_(x->-3) x^2 = 9#