mathop {lim }limits_{x to infty } frac{{(x - | vedclass

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(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{{(x - 1)(2x + 3)}}{{{x^2}}}$.First,expand the numerator: $(x - 1)(2x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{{(x - 1)(2x + 3)}}{{{x^2}}}$.First,expand the numerator: $(x - 1)(2x + 3) = 2x^2 + 3x - 2x - 3 = 2x^2 + x - 3$.Now,substitute this into the limit expression: $\mathop {\lim }\limits_{x 	o \infty } \frac{{2x^2 + x - 3}}{{x^2}}$.Divide each term in the numerator by $x^2$: $\mathop {\lim }\limits_{x 	o \infty } (\frac{{2x^2}}{{x^2}} + \frac{x}{{x^2}} - \frac{3}{{x^2}}) = \mathop {\lim }\limits_{x 	o \infty } (2 + \frac{1}{x} - \frac{3}{{x^2}})$.As $x 	o \infty$,$\frac{1}{x} 	o 0$ and $\frac{3}{{x^2}} 	o 0$.Therefore,the limit is $2 + 0 - 0 = 2$.

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

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