# limit - more about division

• April 20th 2006, 11:57 AM
tnkfub
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
• April 20th 2006, 01:21 PM
Jameson
$\lim_{x\rightarrow{1}}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$

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

Now would be a good time to use L'Hopital's rule, if you have learned this. Have you?
• April 21st 2006, 08:59 PM
earboth
Quote:

Originally Posted by tnkfub
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

Hello,

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:

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

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

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

Greetings

EB
• April 22nd 2006, 05:04 PM
ThePerfectHacker
Quote:

Originally Posted by tnkfub
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

You have,
$\frac{\frac{1}{\sqrt{x}}-1}{x-1}$
You can express as,
$\frac{\frac{\sqrt{x}}{x}-1}{x-1}=\frac{\sqrt{x}-x}{x(x-1)}$
Thus,
$\frac{\sqrt{x}-x}{x(x-1)}\cdot \frac{\sqrt{x}+x}{\sqrt{x}+x}$
Thus,
$\frac{x-x^2}{x(x-1)(\sqrt{x}+x)}$
Thus,
$\frac{-x(x-1)}{x(x-1)(\sqrt{x}+x)}$
Thus,
$-\frac{1}{\sqrt{x}+x}$
As, $x\to 1$ you have,
$-\frac{1}{\sqrt{1}+1}=-\frac{1}{2}$
• April 22nd 2006, 05:24 PM
ThePerfectHacker
Quote:

Originally Posted by earboth
Hello,

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:

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

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

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

Greetings

EB

Can you prove that such a substitution is valid?
Not that I am arguing I am simply interested.
• April 22nd 2006, 10:37 PM
earboth
Quote:

Originally Posted by ThePerfectHacker
Can you prove that such a substitution is valid?
Not that I am arguing I am simply interested.

Hello,

honestly speaking: No.

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

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
• April 23rd 2006, 06:30 AM
ThePerfectHacker
Quote:

Originally Posted by earboth
Hello,

honestly speaking: No.

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

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

Not arguing with what you are saying, and I would use it if I was taking an exam without knowing its proof. I was just always curious about why we can do that. Seen it used many times.