Choose any epsilon > 0ε>0 and evaluate the difference:
abs (sqrt(x+1) -2)∣∣√x+1−2∣∣
we have:
abs (sqrt(x+1) -2) = abs ( (sqrt(x+1) -2) (sqrt(x+1) +2))/ (sqrt(x+1) +2)∣∣√x+1−2∣∣=∣∣(√x+1−2)(√x+1+2)∣∣√x+1+2
abs (sqrt(x+1) -2) = abs ( (x+1) -4 )/ (sqrt(x+1) +2)∣∣√x+1−2∣∣=|(x+1)−4|√x+1+2
abs (sqrt(x+1) -2) = abs ( x-3)/ (sqrt(x+1) +2)∣∣√x+1−2∣∣=|x−3|√x+1+2
Now consider that:
1/(sqrt(x+1) +2)1√x+1+2
is a strictly decreasing function, so for x in (3-delta, 3+delta)x∈(3−δ,3+δ):
1/(sqrt(x+1) +2) <= 1/(sqrt(3-delta+1) +2)= 1/(sqrt(4-delta) +2)1√x+1+2≤1√3−δ+1+2=1√4−δ+2
and clearly:
abs (x- 3) <= delta|x−3|≤δ
so:
abs (sqrt(x+1) -2) <= delta/(sqrt(4-delta) +2)∣∣√x+1−2∣∣≤δ√4−δ+2
So, if we choose delta_epsilonδε such that:
delta_epsilon/(sqrt(4-delta_epsilon) +2) < epsilonδε√4−δε+2<ε
we have:
abs (x-3) < delta_epsilon => abs (sqrt(x+1) -2) < epsilon|x−3|<δε⇒∣∣√x+1−2∣∣<ε