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 (√(x2+4x+1) - x)*(√(x2+4x+1) + x)/(√(x2+4x+1) + x) =
(4x+1)/(√(x2+4x+1) + x)
For x > 0,
(4x+1)/(√(x2+4x+4) + x) < (4x+1)/(√(x2+4x+1) + x) < (4x+1)/(√(x2+2x+1) + x) , or
(4x+1)/(2x+2) < (4x+1)/(√(x2+4x+1) + x) < (4x+1)/(2x+1)
2 - 3/(2x+2) < (4x+1)/(√(x2+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)/(√(x2+4x+1) + x) = 2
lim x -> ∞ √(x2+4x+1) - x = 2
Michelle M.
thank you so much !!01/23/21