Answer
4
Work Step by Step
We want to find $\lim\limits_{x \to 1}\dfrac{x-1}{\sqrt{x+3}-2}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 1 is zero, and if we try to substitute $x=1$ directly, we get zero in both the numerator and the denominator.
But note that $\dfrac{x-1}{\sqrt{x+3}-2}=\left(\dfrac{x-1}{\sqrt{x+3}-2}\right)\left(\dfrac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\right)=\dfrac{(x-1)(\sqrt{x+3}+2)}{x+3-4}=\dfrac{(x-1)(\sqrt{x+3}+2)}{x-1}=\sqrt{x+3}+2$
for all $x \neq 1.$
Thus $\lim\limits_{x \to 1}\dfrac{x-1}{\sqrt{x+3}-2}=\lim\limits_{x \to 1}\sqrt{x+3}+2=\sqrt{1+3}+2=\sqrt{4}+2=2+2=4.$
