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प्रश्न
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
उत्तर
\[\lim_{x \to 3} \left[ \frac{\sqrt{x + 3} - \sqrt{16}}{x^2 - 9} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 3} \left[ \frac{\left( \sqrt{x + 3} - \sqrt{6} \right)\left( \sqrt{x + 3} + \sqrt{6} \right)}{\left( x^2 - 9 \right)\left( \sqrt{x + 3} + \sqrt{6} \right)} \right]\]
= \[\lim_{x \to 3} \left[ \frac{\left( x + 3 - 6 \right)}{\left( x - 3 \right)\left( x + 3 \right)\left( \sqrt{x + 3} + \sqrt{6} \right)} \right]\]
= \[\frac{1}{6 \times 2\sqrt{6}}\]
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