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
$\displaystyle \lim_{x\rightarrow{1}}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$
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?
Hello,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
You have,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,
$\displaystyle -\frac{1}{\sqrt{1}+1}=-\frac{1}{2}$
Hello,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