Question

$\text{If}L=\underset{x\to \infty }{lim}\frac{1^2}{1+x^3}+\frac{2^2}{2+x^3}+...+\frac{x^2}{x+x^3},\text{then the value of}18L\text{is}$

Answer 1

$\text{Let}S=\frac{1^2}{1+x^3}+\frac{2^2}{2+x^3}+...+\frac{x^2}{x+x^3}S<\frac{1^2}{1+x^3}+\frac{2^2}{1+x^3}+...+\frac{x^2}{1+x^3}=\frac{1}{1+x^3}(1^2+2^2+...+x^2)=\frac{x(x+1)(2x+1)}{6(1+x^3)}S>\frac{1^2}{x+x^3}+\frac{2^2}{x+x^3}+...+\frac{x^2}{x+x^3}=\frac{1}{x+x^3}(1^2+2^2+...+x^2)=\frac{x(x+1)(2x+1)}{6(x+x^3)}$

$\text{Now, using Sandwhich theorem, we have}$
$\frac{x(x+1)(2x+1)}{6(x+x^3)}<S<\frac{x(x+1)(2x+1)}{6(1+x^3)}\Rightarrow \underset{x\to \infty }{lim}\frac{2x^3+3x^2+x}{6(x+x^3)}<\underset{x\to \infty }{lim}S<\underset{x\to \infty }{lim}\frac{2x^3+3x^2+x}{6(1+x^3)}\Rightarrow \underset{x\to \infty }{lim}\frac{2+\frac{3}{x}+\frac{1}{x^2}}{6(\frac{1}{x^2}+1)}<\underset{x\to \infty }{lim}S<\underset{x\to \infty }{lim}\frac{2+\frac{3}{x}+\frac{1}{x^2}}{6(\frac{1}{x^3}+1)}\Rightarrow \frac{2}{6}<\underset{x\to \infty }{lim}S<\frac{2}{6}\Rightarrow \frac{1}{3}<\underset{x\to \infty }{lim}S<\frac{1}{3}\therefore L=\underset{x\to \infty }{lim}S=\frac{1}{3}\Rightarrow 18L=6$