tính : $\lim(\sqrt{n+10}-\sqrt{n})$ quang cuong giúp 24/10/2021 Bởi Samantha tính : $\lim(\sqrt{n+10}-\sqrt{n})$ quang cuong giúp
`lim (\sqrt[]n + 10 – \sqrt[]n )` `= lim (\frac{n+10-n}{\sqrt[]{n + 10} + \sqrt[]n})` `= lim (\frac{10}{\sqrt[]{n}.(\sqrt[]{1+\frac{10}{n} } + 1) })` `Vì lim \sqrt[]n = +∞` ` lim (\sqrt[]{1+\frac{10}{n} } + 1) = √1 + 1 = 2 > 0` `=> lim (\sqrt[]{n}.(\sqrt[]{1+\frac{10}{n} } + 1)) = +∞` `=> lim (\frac{10}{\sqrt[]{n}.(\sqrt[]{1+\frac{10}{n} } + 1) }) = 0` `Vậy lim (\sqrt[]n + 10 – \sqrt[]n ) = 0`. Bình luận
+ Ta có: $\lim (\sqrt {n + 10} – \sqrt {n}) $ $= \lim \frac{(\sqrt {n + 10} – \sqrt {n})(\sqrt {n + 10} + \sqrt {n})}{\sqrt {n + 10} + \sqrt {n}}$ $= \lim \frac{n + 10 – n}{\sqrt {n + 10} + \sqrt {n}}$ $= \lim \frac{10}{\sqrt {n + 10} + \sqrt {n}}$ $= 0$. (Do $\lim (\sqrt {n + 10} + \sqrt {n}) = +∞$). XIN HAY NHẤT CHÚC EM HỌC TỐT Bình luận
`lim (\sqrt[]n + 10 – \sqrt[]n )`
`= lim (\frac{n+10-n}{\sqrt[]{n + 10} + \sqrt[]n})`
`= lim (\frac{10}{\sqrt[]{n}.(\sqrt[]{1+\frac{10}{n} } + 1) })`
`Vì lim \sqrt[]n = +∞`
` lim (\sqrt[]{1+\frac{10}{n} } + 1) = √1 + 1 = 2 > 0`
`=> lim (\sqrt[]{n}.(\sqrt[]{1+\frac{10}{n} } + 1)) = +∞`
`=> lim (\frac{10}{\sqrt[]{n}.(\sqrt[]{1+\frac{10}{n} } + 1) }) = 0`
`Vậy lim (\sqrt[]n + 10 – \sqrt[]n ) = 0`.
+ Ta có:
$\lim (\sqrt {n + 10} – \sqrt {n}) $
$= \lim \frac{(\sqrt {n + 10} – \sqrt {n})(\sqrt {n + 10} + \sqrt {n})}{\sqrt {n + 10} + \sqrt {n}}$
$= \lim \frac{n + 10 – n}{\sqrt {n + 10} + \sqrt {n}}$
$= \lim \frac{10}{\sqrt {n + 10} + \sqrt {n}}$
$= 0$. (Do $\lim (\sqrt {n + 10} + \sqrt {n}) = +∞$).
XIN HAY NHẤT
CHÚC EM HỌC TỐT