Answer
$\lim\limits_{x \to 1}\frac{x-1}{\sqrt {x+3} - 2} = 4$
Work Step by Step
$\frac{x-1}{\sqrt {x+3} - 2} = \frac{(x-1)(\sqrt {x+3} + 2)}{(\sqrt {x+3} - 2)(\sqrt {x+3} + 2)} = \frac{(x-1)(\sqrt {x+3} + 2)}{{x+3}-4} = \frac{(x-1)(\sqrt {x+3} + 2)}{(x-1)} = (\sqrt {x+3} + 2)$
Now,
$\lim\limits_{x \to 1}\frac{x-1}{\sqrt {x+3} - 2} =\lim\limits_{x \to 1}(\sqrt {x+3} + 2) = 4$
