You have $\displaystyle \lim_{x\to 3}\frac{x- 3}{\sqrt{x+1}- 2}$ and are asking how that becomes
$\displaystyle 2\lim_{x\to 3}\sqrt{x+1}$
Easy answer- only by accident! The limits are the same but in general
$\displaystyle \frac{x- 3}{\sqrt{x+1}- 2}$ is NOT equal to 2\sqrt{x+ 1}.
For example, if x= 8, the first is equal to $\displaystyle \frac{8- 3}{\sqrt{8+1}- 2}= 5$ while the second is $\displaystyle 2\sqrt{8+ 1}= 6$
What is true is that you can "rationalize the denominator to get
$\displaystyle \frac{x- 3}{\sqrt{x+1}- 2}\fract{\sqrt{x+ 1}+ 2}{\sqrt{x+ 1}+ 2}= \frac{(x- 3)(\sqrt{x+ 1}+ 2)}{x- 3}= \sqrt{x+1}+ 2$
Now, it happens that both [itex]2\sqrt{x+1}[/itex] and [itex]\sqrt{x+1}+ 2[/itex] 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
In 'normal' circustances it shold be...
$\displaystyle \displaystyle \frac{x-3}{\sqrt{x+1}-2}= 2+\sqrt{x+1}$
In the jpg file however is...
$\displaystyle \displaystyle \frac{x-3}{\sqrt{x+1}-2}= 2\ \sqrt{x+1}$
May be that the '+' between 2 and the square root has been 'forgotten'...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$