# What is the limit of (x + (sqrt(x^2 + 2x)) as x goes to negative infinity?

Refer to explanation

#### Explanation:

Divide and multiply with conjugate $x - \sqrt{{x}^{2} + 2 x}$

Hence we have

lim_(x->-oo)((x+sqrt(x^2+2x))*(x-sqrt(x^2+2x)))/(x-sqrt(x^2+2x))= lim_(x->-oo)(x^2-(x^2+2x))/((x-sqrt(x^2+2x)))= lim_(x->-oo)(2x)/(x(sqrt(1+2/x)-1))= lim_(x->-oo)2/(sqrt(1+2/x)-1)= lim_(x->-oo) (-2)/(1+sqrt(1+2/x))=-1

Remarks

Notice that since $x$ is tending to $- \infty$ then $x = - \sqrt{{x}^{2}}$. So when you let $x$ enter the square root in the denominator the minus that was originally there becomes a plus due to cancellation.