1. ## simple limit...

Hi guys,

I have this limit, I fail to understand why the first stage is correct (the transformation of the original function to the state of it after the first stage)

any explanation will be appreciated...thanks !

2. They multiplied top and bottom by the bottom's conjugate...

3. You have $\lim_{x\to 3}\frac{x- 3}{\sqrt{x+1}- 2}$ and are asking how that becomes
$2\lim_{x\to 3}\sqrt{x+1}$

Easy answer- only by accident! The limits are the same but in general
$\frac{x- 3}{\sqrt{x+1}- 2}$ is NOT equal to 2\sqrt{x+ 1}.
For example, if x= 8, the first is equal to $\frac{8- 3}{\sqrt{8+1}- 2}= 5$ while the second is $2\sqrt{8+ 1}= 6$

What is true is that you can "rationalize the denominator to get
$\frac{x- 3}{\sqrt{x+1}- 2}\fract{\sqrt{x+ 1}+ 2}{\sqrt{x+ 1}+ 2}= \frac{(x- 3)(\sqrt{x+ 1}+ 2)}{x- 3}= \sqrt{x+1}+ 2$

Now, it happens that both $2\sqrt{x+1}$ and $\sqrt{x+1}+ 2$ are equal to 4 when x= 3, so the limits are equal, but that is the only value of x for which they are the same

4. In 'normal' circustances it shold be...

$\displaystyle \frac{x-3}{\sqrt{x+1}-2}= 2+\sqrt{x+1}$

In the jpg file however is...

$\displaystyle \frac{x-3}{\sqrt{x+1}-2}= 2\ \sqrt{x+1}$

May be that the '+' between 2 and the square root has been 'forgotten'...

Kind regards

$\chi$ $\sigma$