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प्रश्न
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
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उत्तर
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to \sqrt{3}} \left[ \frac{\left( x^2 \right)^2 - \left( 3 \right)^2}{x^2 + 5\sqrt{3}x - \sqrt{3}x - 15} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x^2 - 3 \right)\left( x^2 + 3 \right)}{x\left( x + 5\sqrt{3} \right) - \sqrt{3}\left( x + 5\sqrt{3} \right)} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left\{ x^2 - \left( \sqrt{3} \right)^2 \right\}\left( x^2 + 3 \right)}{\left( x - \sqrt{3} \right)\left( x + 5\sqrt{3} \right)} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right)\left( x^2 + 3 \right)}{\left( x - \sqrt{3} \right)\left( x + 5\sqrt{3} \right)} \right]\]
\[ = \frac{\left( \sqrt{3} + \sqrt{3} \right)\left( 3 + 3 \right)}{\sqrt{3} + 5\sqrt{3}}\]
\[ = \frac{2\sqrt{3} \times 6}{6\sqrt{3}}\]
\[ = 2\]
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