mathop {lim }limits_{n to infty } frac{{n{{(2 | vedclass

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(C) We need to evaluate the limit: $\mathop {\lim }\limits_{n 	o \infty } \frac{{n{{(2n + 1)}^2}}}{{(n + 2)({n^2} + 3n - 1)}}$.Expanding the numerato

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{n 	o \infty } \frac{{n{{(2n + 1)}^2}}}{{(n + 2)({n^2} + 3n - 1)}}$.Expanding the numerator: $n(2n+1)^2 = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.Expanding the denominator: $(n+2)(n^2 + 3n - 1) = n^3 + 3n^2 - n + 2n^2 + 6n - 2 = n^3 + 5n^2 + 5n - 2$.Now,the limit becomes: $\mathop {\lim }\limits_{n 	o \infty } \frac{{4n^3 + 4n^2 + n}}{{n^3 + 5n^2 + 5n - 2}}$.Divide both the numerator and the denominator by $n^3$:$= \mathop {\lim }\limits_{n 	o \infty } \frac{{4 + \frac{4}{n} + \frac{1}{n^2}}}{{1 + \frac{5}{n} + \frac{5}{n^2} - \frac{2}{n^3}}}$.As $n 	o \infty$,all terms with $n$ in the denominator approach $0$:$= \frac{{4 + 0 + 0}}{{1 + 0 + 0 - 0}} = 4$.

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

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