Question
$\mathop {\lim }\limits_{x \to \infty } \frac{{(x + 1)(3x + 4)}}{{{x^2}(x - 8)}}$ का मान है
$ = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{1}{x}\frac{{\left( {1 + \frac{1}{x}} \right)\,\left( {3 + \frac{4}{x}} \right)}}{{\left( {1 - \frac{8}{x}} \right)}}} \right] = 0$.
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