Question

$\underset{x\to \infty }{lim}\frac{\sqrt{3}{x}^5+x^2+13}{x^4+7x^2-\sqrt{7}}$

• A. $0$
• B. $2$
• C. Infinite
• D. None of these

Answer 1

The correct option is A $0$

$L={lim}_{x\to \infty }\frac{\sqrt{3}{x}^5+x^2+13}{x^4+7x^2-\sqrt{7}}$

Dividing numerator and denominator by highest power of $x$ i.e., $x^5$

${lim}_{x\to \infty }\frac{\sqrt{3}+\frac{1}{x^3}+\frac{13}{x^5}}{\frac{1}{x}+\frac{7}{x^3}-\frac{\sqrt{7}}{x^5}}=\frac{1}{\infty }=0\therefore L=0$