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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
\[\lim_{x \to 3} \left[ \frac{1}{x - 3} - \frac{2}{x^2 - 4x + 3} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{1}{x - 3} - \frac{2}{x^2 - 3x - x + 3} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{1}{x - 3} - \frac{2}{x\left( x - 3 \right) - 1\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{1}{x - 3} - \frac{2}{\left( x - 1 \right)\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{x - 1 - 2}{\left( x - 3 \right)\left( x - 1 \right)} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{1}{x - 1} \right]\]
\[ = \frac{1}{3 - 1}\]
\[ = \frac{1}{2}\]
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