lim:x tiến tới 1 (căn bậc 3 của x+7 )-2 tất cả trên x-1 28/10/2021 Bởi Ariana lim:x tiến tới 1 (căn bậc 3 của x+7 )-2 tất cả trên x-1
$\lim\limits_{x\to 1}\dfrac{ \sqrt[3]{x+7}-2}{x-1}$ $=\lim\limits_{x\to 1}\dfrac{x+7-8}{(x-1)(\sqrt[3]{x+7}^2+2\sqrt[3]{x+7}+4)}$ $=\lim\limits_{x\to 1}\dfrac{1}{\sqrt[3]{x+7}^2+2\sqrt[3]{x+7}+4}$ $=\dfrac{1}{\sqrt[3]{1+7}^2+2\sqrt[3]{1+7}+4}$ $=\dfrac{1}{12}$ Bình luận
`lim_{x -> 1}` $\dfrac{\sqrt[3]{x + 7} – 2}{x – 1}$ `= lim_{x -> 1}` $\dfrac{x + 7 – 8}{(x – 1)[(\sqrt[3]{x + 7})^{2} + 2.\sqrt[3]{x + 7} + 8]}$ `= lim_{x -> 1}` $\dfrac{1}{(\sqrt[3]{x + 7})^{2} + 2.\sqrt[3]{x + 7} + 8}$ `= 1/((\sqrt[3]{1 + 7})^{2} + 2.\sqrt[3]{1 + 7} + 8)` `= 1/16` Bình luận
$\lim\limits_{x\to 1}\dfrac{ \sqrt[3]{x+7}-2}{x-1}$
$=\lim\limits_{x\to 1}\dfrac{x+7-8}{(x-1)(\sqrt[3]{x+7}^2+2\sqrt[3]{x+7}+4)}$
$=\lim\limits_{x\to 1}\dfrac{1}{\sqrt[3]{x+7}^2+2\sqrt[3]{x+7}+4}$
$=\dfrac{1}{\sqrt[3]{1+7}^2+2\sqrt[3]{1+7}+4}$
$=\dfrac{1}{12}$
`lim_{x -> 1}` $\dfrac{\sqrt[3]{x + 7} – 2}{x – 1}$
`= lim_{x -> 1}` $\dfrac{x + 7 – 8}{(x – 1)[(\sqrt[3]{x + 7})^{2} + 2.\sqrt[3]{x + 7} + 8]}$
`= lim_{x -> 1}` $\dfrac{1}{(\sqrt[3]{x + 7})^{2} + 2.\sqrt[3]{x + 7} + 8}$
`= 1/((\sqrt[3]{1 + 7})^{2} + 2.\sqrt[3]{1 + 7} + 8)`
`= 1/16`