Toán $\lim_{x \to -\infty} \sqrt{x^2 – x + 1} + x – 2 $ 05/09/2021 By Sadie $\lim_{x \to -\infty} \sqrt{x^2 – x + 1} + x – 2 $
Đáp án: `-3/2` Giải thích các bước giải: `lim_{n->-\infty} \sqrt{n²-n+1}+n-2` `=lim_{n->-\infty} (\sqrt{n²-n+1}+n) -lim_{n->-\infty} 2` `= lim_{n->-\infty} \frac{n² -n+1-n²}{\sqrt{n²-n+1}-n} -lim_{n->-\infty}2` `= lim_{n->-\infty} \frac{-n+1}{\sqrt{n²-n+1}-n} -lim_{n->-\infty}2` `= \frac{-1+\frac{1}{n}}{-\sqrt{1-1/n +\frac{1}{n²}-1}} -2` `= 1/2 -2 =-3/2` Trả lời
Đáp án: $\lim\limits_{x\to-\infty}\sqrt{x^2-x+1}+x-2=-\dfrac32$ Giải thích các bước giải: $\lim\limits_{x\to-\infty}\sqrt{x^2-x+1}+x-2$ $=\lim\limits_{x\to-\infty}\dfrac{(\sqrt{x^2-x+1}+x-2)(\sqrt{x^2-x+1}-x+2)}{\sqrt{x^2-x+1}-x+2}$ $=\lim\limits_{x\to-\infty}\dfrac{x^2-x+1-x^2+4x-4}{\sqrt{x^2-x+1}-x+2}$ $=\lim\limits_{x\to -\infty}\dfrac{3x-3}{\sqrt{x^2\bigg(1-\dfrac1x+\dfrac1{x^2}\bigg)}-x+2}$ $=\lim\limits_{x\to -\infty}\dfrac{x\bigg(3-\dfrac3x\bigg)}{-x\sqrt{1-\dfrac1x+\dfrac1{x^2}}-x+2}$ $=\lim\limits_{x\to -\infty}\dfrac{x\bigg(3-\dfrac3x\bigg)}{x\bigg(-\sqrt{1-\dfrac1x+\dfrac1{x^2}}-1+\dfrac2x\bigg)}$ $=\lim\limits_{x\to -\infty}\dfrac{3-\dfrac3x}{-\sqrt{1-\dfrac1x+\dfrac1{x^2}}-1+\dfrac2x}$ $=\dfrac{3-0}{-\sqrt{1-0+0}-1+0}=-\dfrac32$ Trả lời
Đáp án: `-3/2`
Giải thích các bước giải:
`lim_{n->-\infty} \sqrt{n²-n+1}+n-2`
`=lim_{n->-\infty} (\sqrt{n²-n+1}+n) -lim_{n->-\infty} 2`
`= lim_{n->-\infty} \frac{n² -n+1-n²}{\sqrt{n²-n+1}-n} -lim_{n->-\infty}2`
`= lim_{n->-\infty} \frac{-n+1}{\sqrt{n²-n+1}-n} -lim_{n->-\infty}2`
`= \frac{-1+\frac{1}{n}}{-\sqrt{1-1/n +\frac{1}{n²}-1}} -2`
`= 1/2 -2 =-3/2`
Đáp án:
$\lim\limits_{x\to-\infty}\sqrt{x^2-x+1}+x-2=-\dfrac32$
Giải thích các bước giải:
$\lim\limits_{x\to-\infty}\sqrt{x^2-x+1}+x-2$
$=\lim\limits_{x\to-\infty}\dfrac{(\sqrt{x^2-x+1}+x-2)(\sqrt{x^2-x+1}-x+2)}{\sqrt{x^2-x+1}-x+2}$
$=\lim\limits_{x\to-\infty}\dfrac{x^2-x+1-x^2+4x-4}{\sqrt{x^2-x+1}-x+2}$
$=\lim\limits_{x\to -\infty}\dfrac{3x-3}{\sqrt{x^2\bigg(1-\dfrac1x+\dfrac1{x^2}\bigg)}-x+2}$
$=\lim\limits_{x\to -\infty}\dfrac{x\bigg(3-\dfrac3x\bigg)}{-x\sqrt{1-\dfrac1x+\dfrac1{x^2}}-x+2}$
$=\lim\limits_{x\to -\infty}\dfrac{x\bigg(3-\dfrac3x\bigg)}{x\bigg(-\sqrt{1-\dfrac1x+\dfrac1{x^2}}-1+\dfrac2x\bigg)}$
$=\lim\limits_{x\to -\infty}\dfrac{3-\dfrac3x}{-\sqrt{1-\dfrac1x+\dfrac1{x^2}}-1+\dfrac2x}$
$=\dfrac{3-0}{-\sqrt{1-0+0}-1+0}=-\dfrac32$