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प्रश्न
\[\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\]
उत्तर
\[\lim_{x \to 2} \left[ \frac{\sqrt{3 - x} - 1}{2 - x} \right]\] It is of the form \[\frac{0}{0} .\]Rationalising the numerator:
\[\lim_{x \to 2} \left[ \frac{\left( \sqrt{3 - x} - 1 \right)\left( \sqrt{3 - x} + 1 \right)}{\left( 2 - x \right)\left( \sqrt{3 - x} + 1 \right)} \right]\]
\[= \lim_{x \to 2} \left[ \frac{3 - x - 1}{\left( 2 - x \right)\left( \sqrt{3 - x} + 1 \right)} \right]\]
\[= \lim_{x \to 2} \left[ \frac{\left( 2 - x \right)}{\left( 2 - x \right)\left( \sqrt{3 - x} + 1 \right)} \right]\]
\[= \frac{1}{\sqrt{3 - 2} + 1}\]
\[ = \frac{1}{1 + 1}\]
\[ = \frac{1}{2}\]
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