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प्रश्न
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
उत्तर
\[\text{ Dividing the numerator and the denominator by } x^2 : \]
\[ \lim_{x \to \infty} \frac{\left( \frac{3x - 1}{x} \right) \left( \frac{4x - 2}{x} \right)}{\left( \frac{x + 8}{x} \right) \left( \frac{x - 1}{x} \right)}\]
\[ = \lim_{x \to \infty} \left[ \frac{\left( 3 - \frac{1}{x} \right) \left( 4 - \frac{2}{x} \right)}{\left( 1 + \frac{8}{x} \right) \left( 1 - \frac{1}{x} \right)} \right]\]
\[\text{ When } x \to \infty , \text{ then } \frac{1}{x} \to 0 . \]
\[\frac{3 \times 4}{1}\]
\[ = 12\]
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