Evaluating the limit $\lim_{x\rightarrow 1} \frac{\sqrt{x+3}-2}{x-1}$ The question is :
$$\lim_{x\rightarrow 1} \frac{\sqrt{x+3}-2}{x-1}$$
I know I probably have to do some sort of factorisation of the numerator in order to cancel the denominator, but the surd has me stumped I'm afraid.
 A: If you multiply both the numerator and the denominator by the conjugate of the expression, you'll get:
\begin{align*}
\lim_{x \to 1}\frac{\sqrt{x + 3} -2}{x - 1} \cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2} & = \lim_{x \to 1} \frac{x + 3 - 4}{(x-1)(\sqrt{x + 3} + 2)}\\
& = \lim_{x \to 1} \frac{x-1}{(x-1)(\sqrt{x + 3} + 2)}\\
& = \lim_{x \to 1} \frac{1}{\sqrt{x + 3} + 2}\\
& = \frac{1}{4}
\end{align*}
Like pointed out by André Nicolas, this trick (multiplying by the conjugate) is quite classic.
A: You end up with $\frac{0}{0}$ if you simply substitute $1$ for $x$. So, I would try L'Hôpital's rule. Take the derivative of the numerator and divide by the derivative of the denominator. See what happens after that.  
$$\lim_{x\rightarrow 1} \frac{\sqrt{x+3}-2}{x-1}=\lim_{x\rightarrow 1} \frac{\frac{d}{dx}\sqrt{x+3}-2}{\frac{d}{dx}(x-1)}=\lim_{x\rightarrow 1}\frac{1}{2\sqrt{x+3}} $$  
Then, substituting $x:=1$
$$\frac{1}{2\sqrt{1+3}} = \frac{1}{4} $$
A: Let $x+3 = a^2$ and $2 = b\;,$ Then $(x+3)-2^2 = a^2-b^2$
So $$\displaystyle \lim_{x\rightarrow 1} \frac{a-b}{a^2-b^2} = \lim_{x\rightarrow 1}\frac{1}{(a+b)} = \lim_{x\rightarrow 1}\frac{1}{\left(\sqrt{x+3}+2\right)} = \frac{1}{4}$$
