Question

$\underset{x\to \infty }{lim}\left\{\frac{x^4\sin \left(\frac{1}{x}\right)+x^2}{1+{\left|x\right|}^3}\right\}$ is

• A. $2$
• B. $1$
• C. $-1$
• D. does not exist

Answer 1

The correct option is B $1$

$\underset{x\to \infty }{lim}\frac{x^4\sin \frac{1}{x}+x^2}{1+{\left|x\right|}^3}$

$=\underset{x\to \infty }{lim}\frac{x^4\frac{\sin \frac{1}{x}}{\frac{1}{x}}\times \frac{1}{x}+x^2}{1+{\left|x\right|}^3}$

$=\underset{x\to \infty }{lim}\frac{x^4\times \frac{1}{x}+x^2}{1+{\left|x\right|}^3}$ since $\underset{x\to \infty }{lim}\frac{\sin \frac{1}{x}}{\frac{1}{x}}=1$

$=\underset{x\to \infty }{lim}\frac{x^3+x^2}{1+{\left|x\right|}^3}$

$=\underset{x\to \infty }{lim}\frac{x^3+x^2}{1+x^3}$

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

$=\underset{x\to \infty }{lim}\frac{1+\frac{1}{x}}{1+\frac{1}{x^3}}$

$=\frac{1+0}{1+0}=1$