Question

${lim}_{x\to 3}\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}$

Answer 1

${lim}_{x\to 3}\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}$

Rationalising the denominator

$={lim}_{x\to 3}\frac{x-3}{(\sqrt{x-2}-\sqrt{4-x})}\times \frac{(\sqrt{x-2}+\sqrt{4-x})}{(\sqrt{x-2}+\sqrt{4-x})}$

$={lim}_{x\to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(x-2)-(4-x)}$

$={lim}_{x\to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{x-2-4+x}$

$={lim}_{x\to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}$

$=\frac{1}{2}{lim}_{x\to 3}(\sqrt{x-2}+\sqrt{4-x})$

$=\frac{1}{2}(\sqrt{3-2}+\sqrt{4-3})$

$=\frac{1}{2}(\sqrt{1}+\sqrt{1})$

$=\frac{1}{2}(1+1)=\frac{2}{2}=1$