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Question: \(\lim_{x \rightarrow 3}\frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\) equals...

lim⁡x→3x−3x−2−4−x\lim_{x \rightarrow 3}\frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}} equals

A

1

B

32\frac{3}{2}

C

14\frac{1}{4}

D

None of these

Answer

None of these

Explanation

Solution

lim⁡x→3x−3x−2−4−x=lim⁡x→3(x−3)(x−2+4−x)(x−2)2−(4−x)2\lim_{x \rightarrow 3}\frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}} = \lim_{x \rightarrow 3}\frac{(x - 3)\left( \sqrt{x - 2} + \sqrt{4 - x} \right)}{\left( \sqrt{x - 2} \right)^{2} - \left( \sqrt{4 - x} \right)^{2}}

=lim⁡x→3(x−3)(x−2+4−x)(2x−6)\lim_{x \rightarrow 3}\frac{(x - 3)\left( \sqrt{x - 2} + \sqrt{4 - x} \right)}{(2x - 6)}= lim⁡x→3x−2+4−x2=1+12=1\lim_{x \rightarrow 3}\frac{\sqrt{x - 2} + \sqrt{4 - x}}{2} = \frac{1 + 1}{2} = 1