`a)lim_{x->+oo} (2x-\sqrt{x^2+4x-3})` `ĐK: x < -2-\sqrt{7};x > -2+\sqrt{7}`
`=lim_{x->+oo} [x(2-\sqrt{1+4/x -3/[x^2]}]`
`=+oo`
`b)lim_{x->+oo} (\sqrt{4x^2-3x+1}-2x)`
`=lim_{x->+oo} [4x^2-3x+1-4x^2]/[\sqrt{4x^2-3x+1}+2x]`
`=lim_{x->+oo} [-3x+1]/[\sqrt{4x^2-3x+1}+2x]`
`=lim_{x->+oo} [-3+1/x]/[\sqrt{4-3/x+1/[x^2]}+2]`
`=-3/4`
Đúng 2
Bình luận (0)