How do you prove that the limit of sqrt(x+1) = 2 x+1=2 as x approaches 3 using the epsilon delta proof?

1 Answer
Feb 22, 2017

See explanation below

Explanation:

Choose any epsilon > 0ε>0 and evaluate the difference:

abs (sqrt(x+1) -2)x+12

we have:

abs (sqrt(x+1) -2) = abs ( (sqrt(x+1) -2) (sqrt(x+1) +2))/ (sqrt(x+1) +2)x+12=(x+12)(x+1+2)x+1+2

abs (sqrt(x+1) -2) = abs ( (x+1) -4 )/ (sqrt(x+1) +2)x+12=|(x+1)4|x+1+2

abs (sqrt(x+1) -2) = abs ( x-3)/ (sqrt(x+1) +2)x+12=|x3|x+1+2

Now consider that:

1/(sqrt(x+1) +2)1x+1+2

is a strictly decreasing function, so for x in (3-delta, 3+delta)x(3δ,3+δ):

1/(sqrt(x+1) +2) <= 1/(sqrt(3-delta+1) +2)= 1/(sqrt(4-delta) +2)1x+1+213δ+1+2=14δ+2

and clearly:

abs (x- 3) <= delta|x3|δ

so:

abs (sqrt(x+1) -2) <= delta/(sqrt(4-delta) +2)x+12δ4δ+2

So, if we choose delta_epsilonδε such that:

delta_epsilon/(sqrt(4-delta_epsilon) +2) < epsilonδε4δε+2<ε

we have:

abs (x-3) < delta_epsilon => abs (sqrt(x+1) -2) < epsilon|x3|<δεx+12<ε