Question

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

Answer 1

Refer to explanation

Explanation:

Divide and multiply with conjugate `x-sqrt(x^2+2x)`

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 `-oo` 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.