How do you prove that the limit of 5x^2 =55x2=5 as x approaches 1 using the epsilon delta proof?

1 Answer
Jul 27, 2018

Evaluate the difference:

abs(5x^2-5) = 5abs(x^2-1)5x25=5x21

abs(5x^2-5) = 5abs(x-1)abs(x+1)5x25=5|x1||x+1|

Now for any epsilon > 0ε>0 choose delta_epsilon < min(1,epsilon/15).

As delta_epsilon < 1 then for x in (1-delta_epsilon, 1+delta_epsilon) we have that:

0 < x < 2

and:

1 < 1+x < 3

Then abs(1+x) = (1+x) < 3 and so:

abs(5x^2-5) = 5abs(x-1)abs(x+1) < 15abs(x-1)

On the other hand, as delta_epsilon < epsilon/15 then for x in (1-delta_epsilon, 1+delta_epsilon) we have that:

abs(x-1) < delta_epsilon < epsilon/15

So:

abs(5x^2-5) < 15abs(x-1)

abs(5x^2-5) < 15epsilon/15

abs(5x^2-5) < epsilon

which proves the limit.