mathop {lim }limits_{x to infty } frac{{3{x^2 | vedclass

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Question

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{3{x^2} + 2x - 1}}{{2{x^2} - 3x - 3}}$,divide the numerator and the denominato

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{3{x^2} + 2x - 1}}{{2{x^2} - 3x - 3}}$,divide the numerator and the denominator by the highest power of $x$,which is $x^2$.$\mathop {\lim }\limits_{x 	o \infty } \frac{{3 + \frac{2}{x} - \frac{1}{{{x^2}}}}}{{2 - \frac{3}{x} - \frac{3}{{{x^2}}}}}$As $x 	o \infty$,the terms $\frac{2}{x}$,$\frac{1}{{{x^2}}}$,$\frac{3}{x}$,and $\frac{3}{{{x^2}}}$ approach $0$.Therefore,the limit is $\frac{{3 + 0 - 0}}{{2 - 0 - 0}} = \frac{3}{2}$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} + 2x - 1}}{{2{x^2} - 3x - 3}} = $

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