$\lim\limits_{x\to-\infty}\dfrac{\sqrt{x^2+2x+3}+4x+1}{\sqrt{4x^2+1}+x-2}$ 25/10/2021 Bởi Adalyn $\lim\limits_{x\to-\infty}\dfrac{\sqrt{x^2+2x+3}+4x+1}{\sqrt{4x^2+1}+x-2}$
$x\to -\infty\Rightarrow \sqrt{x^2}=|x|=-x$ $\lim\limits_{x\to -\infty}\dfrac{ \sqrt{x^2+2x+3}+4x+1}{ \sqrt{4x^2+1}+x-2}$ $=\lim\limits_{x\to -\infty} \dfrac{ -\sqrt{1+\dfrac{2}{x}+\dfrac{3}{x^2}} +4+\dfrac{1}{x} }{ -\sqrt{4+\dfrac{1}{x^2}}+1-\dfrac{2}{x} }$ $=\dfrac{-1+4}{-2+1}$ $=-3$ Bình luận
`lim_{x-> – \infty} (\sqrt{x^2 + 2x + 3} + 4x + 1)/(\sqrt{4x^2 + 1} + x – 2)` `= lim_{x -> -\ infty} (-\sqrt{1 + 2/x + 3/(x^2)} + 4 + 1/x)/(-\sqrt{4 + 1/(x^2)} + 1 – 2/x)` `= (4-1)/(1-2)` `= -3` Bình luận
$x\to -\infty\Rightarrow \sqrt{x^2}=|x|=-x$
$\lim\limits_{x\to -\infty}\dfrac{ \sqrt{x^2+2x+3}+4x+1}{ \sqrt{4x^2+1}+x-2}$
$=\lim\limits_{x\to -\infty} \dfrac{ -\sqrt{1+\dfrac{2}{x}+\dfrac{3}{x^2}} +4+\dfrac{1}{x} }{ -\sqrt{4+\dfrac{1}{x^2}}+1-\dfrac{2}{x} }$
$=\dfrac{-1+4}{-2+1}$
$=-3$
`lim_{x-> – \infty} (\sqrt{x^2 + 2x + 3} + 4x + 1)/(\sqrt{4x^2 + 1} + x – 2)`
`= lim_{x -> -\ infty} (-\sqrt{1 + 2/x + 3/(x^2)} + 4 + 1/x)/(-\sqrt{4 + 1/(x^2)} + 1 – 2/x)`
`= (4-1)/(1-2)`
`= -3`