Question

Evaluate the limit of the function as x approaches negative infinity of: f(x) = (x + \sqrt{x^2 + 2x}

Attempt: Multiplying by the conjugate: = (x + \sqrt{x^2 + 2x} * (x - \sqrt{x^2 + 2x} ---------------- 1 ((x - \sqrt{x^2 + 2x}) Simplifying: = 2x / x - \sqrt{x^2 + 2x} Multiplying both top and bottom by 1/x 2x (1/x) ------- x - \sqrt{x^2 + 2x} Simplifying and I get: 2/0 = DNE.. Is this correct?

Answer 1

\[\sqrt{x^2 + 2x} + x = \frac{2x}{\sqrt{x^2 + 2x}-x} = \frac{2}{ -\sqrt{1+2/x^2} - 1}\]

Answer 2

because for x < 0, sqrt(x^2) = -x

Answer 3

and the limit of that as x -> -infty is -1