x*(căn bậc hai(4*x^2 +9) +2*x) tìm giới hạn hàm số khi x tiến dần tới dương vô cực 23/10/2021 Bởi Delilah x*(căn bậc hai(4*x^2 +9) +2*x) tìm giới hạn hàm số khi x tiến dần tới dương vô cực
`lim_(x->+infty)(xsqrt(4x^2+9)+2x)` `=lim_(x->+infty)[xsqrt(x^2(4+9/x^2))+2x]` `=lim_(x->+infty)(x^2sqrt(4+9/x^2)+2x)` `=lim_(x->+infty)[x(xsqrt(4+9/x^2)+2)]` `=lim_(x->+infty){x[x(sqrt(4+9/x^2)+2/x)]}` `=lim_(x->+infty)[x^2(sqrt(4+9/x^2)+2/x)]` `(1)` Ta có: `lim_(x->+infty)x^2=+infty` `lim_(x->+infty)sqrt(4+9/x^2)+2/x=sqrt4=2` `->(1)=+infty.2=+infty` Bình luận
$\lim\limits_{x\to +\infty}x( \sqrt{4x^2+9}+2x)$ $=\lim\limits_{x\to +\infty} x(x\sqrt{4+\dfrac{9}{x^2}} +2x)$ $=\lim\limits_{x\to +\infty} x^2\Big(\sqrt{4+\dfrac{9}{x^2}}+2\Big)$ $=+\infty$ Bình luận
`lim_(x->+infty)(xsqrt(4x^2+9)+2x)`
`=lim_(x->+infty)[xsqrt(x^2(4+9/x^2))+2x]`
`=lim_(x->+infty)(x^2sqrt(4+9/x^2)+2x)`
`=lim_(x->+infty)[x(xsqrt(4+9/x^2)+2)]`
`=lim_(x->+infty){x[x(sqrt(4+9/x^2)+2/x)]}`
`=lim_(x->+infty)[x^2(sqrt(4+9/x^2)+2/x)]` `(1)`
Ta có:
`lim_(x->+infty)x^2=+infty`
`lim_(x->+infty)sqrt(4+9/x^2)+2/x=sqrt4=2`
`->(1)=+infty.2=+infty`
$\lim\limits_{x\to +\infty}x( \sqrt{4x^2+9}+2x)$
$=\lim\limits_{x\to +\infty} x(x\sqrt{4+\dfrac{9}{x^2}} +2x)$
$=\lim\limits_{x\to +\infty} x^2\Big(\sqrt{4+\dfrac{9}{x^2}}+2\Big)$
$=+\infty$