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प्रश्न
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
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उत्तर
\[\lim_{x \to 3} \left[ \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 3} \left[ \frac{x^2 - 3x + 2x - 6}{x^2 \left( x - 3 \right) + 1\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{x\left( x - 3 \right) + 2\left( x - 3 \right)}{\left( x^2 + 1 \right)\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{\left( x + 2 \right)\left( x - 3 \right)}{\left( x^2 + 1 \right)\left( x - 3 \right)} \right]\]
\[ = \frac{3 + 2}{3^2 + 1}\]
\[ = \frac{5}{10}\]
\[ = \frac{1}{2}\]
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