Evaluate the difference:
abs(5x^2-5) = 5abs(x^2-1)∣∣5x2−5∣∣=5∣∣x2−1∣∣
abs(5x^2-5) = 5abs(x-1)abs(x+1)∣∣5x2−5∣∣=5|x−1||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.