<s>[INST] 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.
 [/INST] 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.
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