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.

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.
