Evaluate the difference:
abs(x^2+3-4) = abs (x^2-1)∣∣x2+3−4∣∣=∣∣x2−1∣∣
abs(x^2+3-4) = abs (x-1)abs(x+1)∣∣x2+3−4∣∣=|x−1||x+1|
Let now x=1+xix=1+ξ, then:
abs(x^2+3-4) = abs xi abs ( 2 +xi)∣∣x2+3−4∣∣=|ξ||2+ξ|
Given any number epsilon > 0ε>0 choose:
delta_epsilon < min(1,epsilon/3)
For x in (1-delta_epsilon,1+delta_epsilon) we have that:
abs xi < delta_epsilon
Now, as delta_epsilon < 1 we have that:
abs(2+xi) <= 2+abs xi < 2+1 = 3
and then:
abs(x^2+3-4) = abs xi abs ( 2 +xi) < 3abs(xi)
and because delta_epsilon < epsilon/3 we have that:
abs(x^2+3-4) < 3abs(xi) < 3 * epsilon/3 = epsilon
We can conclude that for every epsilon if we choose delta_epsilon < min(1,epsilon/3) we have that:
x in (1-delta_epsilon,1+delta_epsilon) => abs(x^2+3-4) < epsilon