Đáp án:
`lim_(x->-infty)x(sqrt(9x^2+1)+3x)=0`
Giải thích các bước giải:
`lim_(x->-infty)x(sqrt(9x^2+1)+3x)`
`=lim_(x->-infty)x[sqrt(x^2(9+1/x^2))+3x]`
`=lim_(x->-infty)x(-xsqrt(9+1/x^2)+3x)`
`=lim_(x->-infty)x^2(-sqrt(9+1/x^2)+3)` `(1)`
Ta có:
`lim_(x->-infty)x^2=+infty`
`lim_(x->-infty)(-sqrt(9+1/x^2)+3)=0`
`=>(1)=+infty.0=0`