Question

$\underset{x\to \infty }{lim}\frac{6x^4+7x^3+2x+1}{x^4+1}$= ______

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

Answer: (D)

6

Given, $\underset{x\to \infty }{lim}\frac{6x^4+7x^3+2x+1}{x^4+1}$

Dividing the numerator and denominator of the fraction by $x^4$, we get

$\underset{x\to \infty }{lim}\frac{6+\frac{7}{x}+\frac{2}{x^2}+\frac{1}{x^4}}{1+\frac{1}{x^4}}$

Substituting $x=\infty$ or $\frac{1}{x}=0$, we get

$=\frac{6+7\left(0\right)+2\left(0\right)+1\left(0\right)}{1+1\left(0\right)}=\frac{6}{1}=6$