If the problem asked for the limit of the square root of x^2+4x+4 - x, what would you think the answer was?

I would guess that you would say x+2 - x = 2, so it is always 2, and the limit is 2.

We shall use the standard trick to generate this result.

We multiply the numerator by conjugate/conjugate.

We then get (√(x^{2}+4x+1) - x)*(√(x^{2}+4x+1) + x)/(√(x^{2}+4x+1) + x) =

(4x+1)/(√(x^{2}+4x+1) + x)

For x > 0,

(4x+1)/(√(x^{2}+4x+4) + x) < (4x+1)/(√(x^{2}+4x+1) + x) < (4x+1)/(√(x^{2}+2x+1) + x) , or

(4x+1)/(2x+2) < (4x+1)/(√(x^{2}+4x+1) + x) < (4x+1)/(2x+1)

2 - 3/(2x+2) < (4x+1)/(√(x^{2}+4x+1) + x) < 2 - 1/(2x+1)

lim x -> ∞ 2 - 3/(2x+2) = 2

lim x -> ∞ 2 - 1/(2x+1) = 2

Thus, lim x -> ∞ (4x+1)/(√(x^{2}+4x+1) + x) = 2

lim x -> ∞ √(x^{2}+4x+1) - x = 2

Michelle M.

thank you so much !!01/23/21