How do you prove that the limit of #x^2 - 5x + 6 = 2# as x approaches 1 using the epsilon delta proof?

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Answer 1 · 2017-05-02T23:21:06

See below.

Given any number #epsilon > 0# choose #delta = min (1, epsilon/4)# then we have that:

For any #x# with #0 < abs(x-1) < delta#,

#-1 < x-1 < 1#, so #-4 < x-4 < -2#

thus #abs(x-4) < 4#

and

#abs(x-1) < epsilon/4#

so that:

If #0 < abs(x-1) < delta# then

#abs((x^2-5x+6) -2) = abs(x-4)abs(x-1) < 4abs(x-1) < 4delta = 4 epsilon/4 = epsilon#