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प्रश्न
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
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उत्तर
\[\lim_{x \to 4} \left[ \frac{x^2 - 7x + 12}{x^2 - 3x - 4} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 4} \left[ \frac{x^2 - 3x - 4x + 12}{x^2 - 4x + x - 4} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{x\left( x - 3 \right) - 4\left( x - 3 \right)}{x\left( x - 4 \right) + 1\left( x - 4 \right)} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{\left( x - 4 \right)\left( x - 3 \right)}{\left( x - 4 \right)\left( x + 1 \right)} \right]\]
\[ = \frac{4 - 3}{4 + 1}\]
\[ = \frac{1}{5}\]
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