`lim_(x->-oo) [sqrt(x^2+x+1)+x]=`

Options

A. `(1)/(2)`
B. `(1)/(2)`
C. `1`
D. `-1`

Answer

B

Explanation

We have,
`lim_(xto-oo)(sqrt(x^2-x+1)+x)`
`=lim_(ytooo)(sqrt(y^2-y+1)-y)`where `y=-x`
`=lim_(ytooo)({sqrt(y^2-y+1)-y}{sqrt(y^2+y+1+y)})/({sqrt(y^2+y+1+y)})`
`=lim_(ytooo)(y^2+y+1-y^2)/(sqrt(y^2+y+1)+y)`
`lim_(ytooo)(y+1)/(sqrt(y^2+y+1+y))=lim_(ytooo)(1+(1)/(y))/(sqrt(1+(1)/(y)+(1)/(y^2))+1)=(1)/(2)`
ALTER
`lim_(xto-oo) (sqrt(x^2-x+1+x))`
` =lim_(xto-oo) ({sqrt(x^2-x+1)+x}{sqrt(x^2-x+1)-x})/({sqrt(x^2-x)+1+x})`
`=lim_(xtooo) (x^2-x+1-x^2)/({sqrt(x^2-x+1)-x})`
`=lim_(xtooo) (-x+1)/(sqrt(x^2-x+1)-x)`
`=lim_(xtooo) ((-x)/(|x|)+(1)/(|x|))/((sqrt(x^2-x+1))/(|x|)-(1)/(|x|))" "["Dividing N' and D' by" |x|]`
`=lim_(xto-oo) ((-x)/(-x)-(1)/(x))/(sqrt(x^2/(x^2)-(x)/(x^2)+(1)/(x^2))+(x)/(x))[because sqrt(x^2)=|x|=-x " as "x lt 0]`
`=lim_(xto-oo) (1-(1)/(x))/(sqrt(1-(1)/(x)+(1)/(x^2))+1)=(1)/(2)`