Question

Question #458e4

Answer 1

`lim_(x->-oo) (sqrt(1+x+x^2)+x)/3 = -1/6`

Explanation:

`lim_(x->-oo) (sqrt(1+x+x^2)+x)/3 = lim_(x->-oo) (absxsqrt(1/x^2+1/x+1)+x)/3`

For `x < 0` we have `abs x = -x`, so:

`lim_(x->-oo) (sqrt(1+x+x^2)+x)/3 = lim_(x->-oo) (x(1- sqrt(1/x^2+1/x+1)))/3`

Rationalize the numerator:

`lim_(x->-oo) (sqrt(1+x+x^2)+x)/3 = lim_(x->-oo) (x(1- sqrt(1/x^2+1/x+1))(1+ sqrt(1/x^2+1/x+1)))/(3(1+ sqrt(1/x^2+1/x+1))`

`lim_(x->-oo) (sqrt(1+x+x^2)+x)/3 = lim_(x->-oo) (x(1- 1/x^2-1/x-1))/(3(1+ sqrt(1/x^2+1/x+1))`

Simplify:

`lim_(x->-oo) (sqrt(1+x+x^2)+x)/3 = lim_(x->-oo) -( 1/x+1)/(3(1+ sqrt(1/x^2+1/x+1))`

and now both numerator and denominator are finite:

`lim_(x->-oo) (sqrt(1+x+x^2)+x)/3 = -1/6`

graph{(sqrt(1+x+x^2)+x)/3 [-10, 1, -0.5, 0.2]}