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Mathematics Question on limits and derivatives

lim⁡n→∞12+22+...+n24n3+6n2−5n+1=\displaystyle\lim_{n\to\infty} \frac{1^{2}+2^{2} +...+n^{2}}{4n^{3}+6n^{2}-5n+1}=

A

16\frac{1}{6}

B

112\frac{1}{12}

C

118\frac{1}{18}

D

14\frac{1}{4}

Answer

112\frac{1}{12}

Explanation

Solution

lim⁡n→∞12+22+...+n24n3+6n2−5n+1\lim_{n\to\infty} \frac{1^{2}+2^{2} +...+n^{2}}{4n^{3}+6n^{2}-5n+1}
=lim⁡n→∞n(+1)(2n+1)64n3+6n2−5n+1= \lim _{n\to \infty } \frac{\frac{n\left(+1\right)\left(2n+1\right)}{6}}{4n^{3}+6n^{2}-5n+1}
=16lim⁡n→∞[2n3+3n2+n4n3+6n2−5n+1]= \frac{1}{6}\lim _{n\to \infty } \left[\frac{2n^{3}+3n^{2}+n}{4n^{3}+6n^{2}-5n+1}\right]
=16lim⁡n→∞[2+3n+1n24+6n−5n2+1n3]=\frac{1}{6}\lim _{n\to \infty } \left[\frac{2+\frac{3}{n}+\frac{1}{n^{2}}}{4+\frac{6}{n}-\frac{5}{n^{2}}+\frac{1}{n^{3}}}\right]
=16[2+0+04+0−0+0=16×24=112]= \frac{1}{6}\left[\frac{2+0+0}{4+0-0+0} =\frac{1}{6} \times\frac{2}{4}=\frac{1}{12}\right]