You have and are asking how that becomes
Easy answer- only by accident! The limits are the same but in general
is NOT equal to 2\sqrt{x+ 1}.
For example, if x= 8, the first is equal to while the second is
What is true is that you can "rationalize the denominator to get
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