Question

$\underset{x\to \infty }{lim}\frac{x(\log x{)}^3}{1+x+x^2}$ equals

• A. $0$
• B. $-1$
• C. $1$
• D. none of these

Answer 1

The correct option is A $0$

$\underset{x\to \infty }{lim}\frac{x(\log x{)}^3}{1+x+x^2}$
=$\underset{x\to \infty }{lim}\frac{(\log x{)}^3+x.3(\log x{)}^2\times \frac{1}{x}}{1+2x}\text{(Using L'Hospital's rule)}$
=$\underset{x\to \infty }{lim}\frac{(\log x{)}^3+3(\log x{)}^2}{1+2x}$
$=\underset{x\to \infty }{lim}\frac{3(\log x{)}^2\times \frac{1}{x}+6\left(\log x\right)\times \frac{1}{x}}{2}${(Using L'Hospital's rule)}
$=\underset{x\to \infty }{lim}\frac{3(\log x{)}^2+6\log x}{2x}\text{(Using L'Hospital's Rule)}$
$=\underset{x\to \infty }{lim}\frac{6\log x\times \frac{1}{x}+\frac{6}{x}}{2}\text{(Using L'Hospital's rule)}$
$=\underset{x\to \infty }{lim}\frac{6\log x+6}{2x}$
$=\underset{x\to \infty }{lim}\frac{6(\frac{1}{x})+0}{2}\text{(Using L'Hospital's rule)}$
$=0$