lim ((1/sqrtx)-1)/(x-1)

x>1

(1-sqrtx/sqrtx)/(x-1)

why does this end up looking like this (1-sqrtx)/(sqrtx(x-1))

i am having a brain fart need this for my exam :D

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- Apr 20th 2006, 11:57 AMtnkfublimit - more about division
lim ((1/sqrtx)-1)/(x-1)

x>1

(1-sqrtx/sqrtx)/(x-1)

why does this end up looking like this (1-sqrtx)/(sqrtx(x-1))

i am having a brain fart need this for my exam :D - Apr 20th 2006, 01:21 PMJameson
$\displaystyle \lim_{x\rightarrow{1}}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$

As you said, this is equivalent to $\displaystyle \lim_{x\rightarrow{1}}\frac{1-\sqrt{x}}{x^\frac{3}{2}-\sqrt{x}}$

Now would be a good time to use L'Hopital's rule, if you have learned this. Have you? - Apr 21st 2006, 08:59 PMearbothQuote:

Originally Posted by**tnkfub**

here is a way to solve this problem without using the l'Hospital rule:

Substitute the x by (1+1/n) and calculate the limit for n approaching infinity:

$\displaystyle \lim_{\csub{n \rightarrow \infty}} {\left( \frac{ \frac{1}{ \sqrt{1+\frac{1}{n}}}-1}{1+\frac{1}{n} - 1} \right)}$

After a few steps of very unpleasant transformation you'll get:

$\displaystyle \lim_{\csub{n \rightarrow \infty}} {\left( \frac{n \cdot \left( n \cdot \sqrt{\frac{n+1}{n}}-n-1 \right)}{n+ 1} \right)}=- \frac{1}{2}$

Greetings

EB - Apr 22nd 2006, 05:04 PMThePerfectHackerQuote:

Originally Posted by**tnkfub**

$\displaystyle \frac{\frac{1}{\sqrt{x}}-1}{x-1}$

You can express as,

$\displaystyle \frac{\frac{\sqrt{x}}{x}-1}{x-1}=\frac{\sqrt{x}-x}{x(x-1)}$

Thus,

$\displaystyle \frac{\sqrt{x}-x}{x(x-1)}\cdot \frac{\sqrt{x}+x}{\sqrt{x}+x}$

Thus,

$\displaystyle \frac{x-x^2}{x(x-1)(\sqrt{x}+x)}$

Thus,

$\displaystyle \frac{-x(x-1)}{x(x-1)(\sqrt{x}+x)}$

Thus,

$\displaystyle -\frac{1}{\sqrt{x}+x}$

As, $\displaystyle x\to 1$ you have,

$\displaystyle -\frac{1}{\sqrt{1}+1}=-\frac{1}{2}$ - Apr 22nd 2006, 05:24 PMThePerfectHackerQuote:

Originally Posted by**earboth**

*prove*that such a substitution is valid?

Not that I am arguing I am simply interested. - Apr 22nd 2006, 10:37 PMearbothQuote:

Originally Posted by**ThePerfectHacker**

honestly speaking: No.

1. When you calculate the derivative you calculate for instance the limit $\displaystyle dx = \lim_{\csub{x \rightarrow x_0}}{(x_0-x)}= \lim_{\csub{n \rightarrow \infty}}{(x_0-(x_0+\frac{1}{n}))$

2. I made an analoguous conclusion to use this method with this problem. I noticed that it worked fine. But of course that isn't a proof.

Greetings

EB - Apr 23rd 2006, 06:30 AMThePerfectHackerQuote:

Originally Posted by**earboth**