Find the value of $\displaystyle \lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2})$.
^include working and explanation needed pls
Hi
$\displaystyle lim_{n \to 4} \sqrt{n}-2 = 0$
$\displaystyle lim_{n \to 4} n - 4 = 0$
L'Hospital:
$\displaystyle lim_{n \to 4} \frac{n-4}{\sqrt{n}-2} = $
$\displaystyle lim_{n \to 4} \frac{1}{(\sqrt{n})' } = lim_{n \to 4} \frac{1}{0.5 * n^{-0.5}}$
$\displaystyle = lim_{n \to 4} n^{0.5}/0.5 =lim_{n \to 4} n^{0.5}*2 = 4^{0.5} * 2 = 4$
$\displaystyle \lim_{n \to 4} \frac{n-4}{\sqrt{n} -2 }$
Multiply by conjugate
$\displaystyle \lim_{n \to 4} \bigg(\frac{n-4}{\sqrt{n} -2 }\bigg)\bigg(\frac{\sqrt{n} + 2 }{\sqrt{n} + 2}\bigg)$
$\displaystyle \lim_{n \to 4} \frac{(n-4)(\sqrt{n} + 2)}{n-4}$
$\displaystyle \lim_{n \to 4} \sqrt{n} +2$
4
With Skalkaz hint (btw it has a typo in it)
$\displaystyle lim \frac{n-4}{\sqrt{n} - 2} = lim \frac{(\sqrt{n})^2 - 2^2}{\sqrt{n} - 2} $
$\displaystyle = lim \frac{(\sqrt{n}-2)(\sqrt{n}+2)}{\sqrt{n}-2} $
$\displaystyle = lim_{n -> 4}\sqrt{n}+2 = \sqrt{4} + 2 = 4$
Regards Rapha