Tìm giới hạn sau: $\mathop {\lim }\limits_{x \to 1} \dfrac{\sqrt{2x-1}-x}{x^2-1}$ 19/10/2021 Bởi Serenity Tìm giới hạn sau: $\mathop {\lim }\limits_{x \to 1} \dfrac{\sqrt{2x-1}-x}{x^2-1}$
Đáp án: `\lim_{x \to 1} \frac{sqrt[2x-1]-x}{x^2-1}=0` Giải thích các bước giải: `\lim_{x \to 1} \frac{sqrt[2x-1]-x}{x^2-1}` `=\lim_{x \to 1} \frac{(sqrt[2x-1]-x)(sqrt[2x-1]+x)}{(x^2-1)(sqrt[2x-1]+x)}` `=\lim_{x \to 1} \frac{2x-1-x^2}{(x+1)(x-1)(sqrt[2x-1]+x)}` `=\lim_{x \to 1} \frac{-(x-1)^2}{(x+1)(x-1)(sqrt[2x-1]+x)}` `=\lim_{x \to 1} \frac{-(x-1)}{(x+1)(sqrt[2x-1]+x)}` `= \frac{-(1-1)}{(1+1)(sqrt[2.1-1]+1)}=0` Vậy `\lim_{x \to 1} \frac{sqrt[2x-1]-x}{x^2-1}=0` Bình luận
$\mathop {\lim }\limits_{x \to 1} \dfrac{\sqrt{2x-1}-x}{x^2-1}$ Ta có: $\mathop {\lim }\limits_{x \to 1} \dfrac{2x-1-x^2}{(x-1)(x+1)(\sqrt{2x-1}+x)}$ $\mathop {\lim }\limits_{x \to 1} \dfrac{-(x-1)}{(x+1)(\sqrt{2x-1}+x)}=0$ Bình luận
Đáp án:
`\lim_{x \to 1} \frac{sqrt[2x-1]-x}{x^2-1}=0`
Giải thích các bước giải:
`\lim_{x \to 1} \frac{sqrt[2x-1]-x}{x^2-1}`
`=\lim_{x \to 1} \frac{(sqrt[2x-1]-x)(sqrt[2x-1]+x)}{(x^2-1)(sqrt[2x-1]+x)}`
`=\lim_{x \to 1} \frac{2x-1-x^2}{(x+1)(x-1)(sqrt[2x-1]+x)}`
`=\lim_{x \to 1} \frac{-(x-1)^2}{(x+1)(x-1)(sqrt[2x-1]+x)}`
`=\lim_{x \to 1} \frac{-(x-1)}{(x+1)(sqrt[2x-1]+x)}`
`= \frac{-(1-1)}{(1+1)(sqrt[2.1-1]+1)}=0`
Vậy `\lim_{x \to 1} \frac{sqrt[2x-1]-x}{x^2-1}=0`
$\mathop {\lim }\limits_{x \to 1} \dfrac{\sqrt{2x-1}-x}{x^2-1}$
Ta có:
$\mathop {\lim }\limits_{x \to 1} \dfrac{2x-1-x^2}{(x-1)(x+1)(\sqrt{2x-1}+x)}$
$\mathop {\lim }\limits_{x \to 1} \dfrac{-(x-1)}{(x+1)(\sqrt{2x-1}+x)}=0$