# fiNd the VaLue of

• Dec 8th 2008, 07:55 PM
mathbuoy
fiNd the VaLue of
Find the value of $\lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2})$.

^include working and explanation needed pls (Wait)
• Dec 8th 2008, 09:27 PM
Skalkaz
Dear mathbuoy,

use this:
$n - 4 = (\sqrt 2)^2 - 2^2$
and
$a^2 - b^2 = (a+b)*(a-b)$
• Dec 8th 2008, 09:29 PM
Rapha
Hi

Quote:

Originally Posted by mathbuoy
Find the value of $\lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2})$.

^include working and explanation needed pls (Wait)

$lim_{n \to 4} \sqrt{n}-2 = 0$
$lim_{n \to 4} n - 4 = 0$

L'Hospital:

$lim_{n \to 4} \frac{n-4}{\sqrt{n}-2} =$
$lim_{n \to 4} \frac{1}{(\sqrt{n})' } = lim_{n \to 4} \frac{1}{0.5 * n^{-0.5}}$

$= lim_{n \to 4} n^{0.5}/0.5 =lim_{n \to 4} n^{0.5}*2 = 4^{0.5} * 2 = 4$
• Dec 8th 2008, 09:47 PM
11rdc11
Quote:

Originally Posted by mathbuoy
Find the value of $\lim(n\rightarrow 4)(\frac{n-4}{\sqrt{n}-2})$.

^include working and explanation needed pls (Wait)

$\lim_{n \to 4} \frac{n-4}{\sqrt{n} -2 }$

Multiply by conjugate

$\lim_{n \to 4} \bigg(\frac{n-4}{\sqrt{n} -2 }\bigg)\bigg(\frac{\sqrt{n} + 2 }{\sqrt{n} + 2}\bigg)$

$\lim_{n \to 4} \frac{(n-4)(\sqrt{n} + 2)}{n-4}$

$\lim_{n \to 4} \sqrt{n} +2$

4
• Dec 8th 2008, 09:48 PM
Rapha
With Skalkaz hint (btw it has a typo in it)

$lim \frac{n-4}{\sqrt{n} - 2} = lim \frac{(\sqrt{n})^2 - 2^2}{\sqrt{n} - 2}$

$= lim \frac{(\sqrt{n}-2)(\sqrt{n}+2)}{\sqrt{n}-2}$

$= lim_{n -> 4}\sqrt{n}+2 = \sqrt{4} + 2 = 4$

Regards Rapha