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प्रश्न
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
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उत्तर
\[\lim_{x \to \sqrt{3}} \left[ \frac{x^2 - 3}{x^2 + 3\sqrt{3}x - 12} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to \sqrt{3}} \left[ \frac{x^2 - \left( \sqrt{3} \right)^2}{x^2 + 4\sqrt{3}x - \sqrt{3}x - 12} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right)}{x\left( x + 4\sqrt{3} \right) - \sqrt{3}\left( x + 4\sqrt{3} \right)} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right)}{\left( x - \sqrt{3} \right)\left( x + 4\sqrt{3} \right)} \right]\]
\[ = \frac{\sqrt{3} + \sqrt{3}}{\sqrt{3} + 4\sqrt{3}}\]
\[ = \frac{2}{5}\]
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