The value of mathop {lim }limits_{x to infty | vedclass

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(D) We are given the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{(x + 1)(3x + 4)}{x^2(x - 8)}$Expand the numerator: $(x + 1)(3x + 4) = 3x^2 +

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(D) We are given the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{(x + 1)(3x + 4)}{x^2(x - 8)}$Expand the numerator: $(x + 1)(3x + 4) = 3x^2 + 4x + 3x + 4 = 3x^2 + 7x + 4$Expand the denominator: $x^2(x - 8) = x^3 - 8x^2$Now,divide both the numerator and the denominator by the highest power of $x$ in the denominator,which is $x^3$:$\mathop {\lim }\limits_{x 	o \infty } \frac{\frac{3x^2}{x^3} + \frac{7x}{x^3} + \frac{4}{x^3}}{\frac{x^3}{x^3} - \frac{8x^2}{x^3}} = \mathop {\lim }\limits_{x 	o \infty } \frac{\frac{3}{x} + \frac{7}{x^2} + \frac{4}{x^3}}{1 - \frac{8}{x}}$As $x 	o \infty$,the terms $\frac{3}{x}, \frac{7}{x^2}, \frac{4}{x^3}$,and $\frac{8}{x}$ all approach $0$.Therefore,the limit is $\frac{0 + 0 + 0}{1 - 0} = \frac{0}{1} = 0$.

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{(x + 1)(3x + 4)}{x^2(x - 8)}$ is equal to

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