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Question
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
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Solution
\[\lim_{x \to 3} \left[ \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}} \right]\] It is of form \[\frac{0}{0} .\]
Rationalising the denominator:
=\[\lim_{x \to 3} \left[ \frac{\left( x - 3 \right)\left( \sqrt{x - 2} + \sqrt{4 - x} \right)}{\left( \sqrt{x - 2} - \sqrt{4 - x} \right)\left( \sqrt{x - 2} + \sqrt{4 - x} \right)} \right]\]
=\[\lim_{x \to 3} \left[ \frac{\left( x - 3 \right)\left( \sqrt{x - 2} + \sqrt{4 - x} \right)}{\left( x - 2 \right) - \left( 4 - x \right)} \right]\]
=\[\lim_{x \to 3} \left[ \frac{\left( x - 3 \right)\left( \sqrt{x - 2} + \sqrt{4 - x} \right)}{2x - 6} \right]\]
=\[\lim_{x \to 3} \left[ \frac{\left( x - 3 \right)\left( \sqrt{x - 2} + \sqrt{4 - x} \right)}{2\left( x - 3 \right)} \right]\]
=\[\frac{\sqrt{3 - 2} + \sqrt{4 - 3}}{2}\]
=\[\frac{\sqrt{1} + \sqrt{1}}{2}\]
=\[\frac{2}{2} = 1\]
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