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प्रश्न
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
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उत्तर
\[\lim_{x \to 3} \left[ \frac{x^2 - 4x + 3}{x^2 - 2x - 3} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 3} \left[ \frac{x^2 - x - 3x + 3}{x^2 - 3x + x - 3} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{\left( x - 3 \right)\left( x - 1 \right)}{\left( x + 1 \right)\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left( \frac{x - 1}{x + 1} \right)\]
\[ = \frac{3 - 1}{3 + 1}\]
\[ = \frac{1}{2}\]
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