Hello, l flipboi l!
Why is it divided by $\displaystyle n^2$?
$\displaystyle \lim_{n\to\infty}\frac{n}{\sqrt{10+n}} \;=\;\lim_{n\to\infty}\dfrac{1}{\sqrt{\dfrac{10}{n ^2}+\dfrac{1}{n}}} \;=\;\infty$
We have the fraction: .$\displaystyle \frac{n}{\sqrt{10+n}}$
$\displaystyle \text{Divide top and bottom by }n\!:$
. . $\displaystyle \dfrac{\dfrac{n}{n}}{\dfrac{\sqrt{10+n}}{n}} \;=\;\dfrac{1}{\dfrac{\sqrt{10+n}}{\sqrt{n^2}}} \;=\;\frac{1}{\sqrt{\dfrac{10+n}{n^2}}} \;=\;\frac{1}{\sqrt{\dfrac{10}{n^2} + \dfrac{n}{n^2}}} \;=\;\frac{1}{\sqrt{\dfrac{10}{n^2} + \dfrac{1}{n}}} $
Got it?