lim _{n rightarrow infty} frac{n(2 n+1)^2}{(n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given limit is $\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)(n^2+3 n-1)}$.Divide the numerator and denominator by the highest power of $n$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given limit is $\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)(n^2+3 n-1)}$.Divide the numerator and denominator by the highest power of $n$,which is $n^3$.Numerator: $n(2n+1)^2 = n(4n^2+4n+1) = 4n^3+4n^2+n$.Denominator: $(n+2)(n^2+3n-1) = n^3+3n^2-n+2n^2+6n-2 = n^3+5n^2+5n-2$.Now,$\lim _{n \rightarrow \infty} \frac{4n^3+4n^2+n}{n^3+5n^2+5n-2} = \lim _{n \rightarrow \infty} \frac{4 + \frac{4}{n} + \frac{1}{n^2}}{1 + \frac{5}{n} + \frac{5}{n^2} - \frac{2}{n^3}}$.As $n \rightarrow \infty$,$\frac{1}{n}, \frac{1}{n^2}, \frac{1}{n^3} \rightarrow 0$.Therefore,the limit is $\frac{4+0+0}{1+0+0-0} = 4$.

$\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)(n^2+3 n-1)}$

Similar Questions

Let $x_n = (2^n + 3^n)^{\frac{1}{2n}}$ for all natural numbers $n$. Then,

$\mathop {\lim }\limits_{x \to 0} {(1 - ax)^{\frac{1}{x}}} = $

The $\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}$ equals

The value of $\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}\,\, - \,\,{{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}}\,$ (where $n \in N$) is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module