Question

$\underset{x\to \infty }{lim}\frac{5x^2}{\sqrt{25x^4+13x^3+14x^2+17x+6}}$ is ______

• A. 1
• B. 1/5
• C. 5/14
• D. None of these

Answer 1

The correct option is A 1

Given, $\underset{x\to \infty }{lim}\frac{5x^2}{\sqrt{25x^4+13x^3+14x^2+17x+6}}$
Dividing the numerator of the fraction by $x^2$ which means dividing the denominator by $x^4$ as it under square root, we get
$\underset{x\to \infty }{lim}\frac{5}{\sqrt{25+\frac{13}{x}+\frac{14}{x^2}+\frac{17}{x^3}+\frac{6}{x^4}}}$
Substituting $x=\infty$ or $\frac{1}{x}=0$, we get

$=\frac{5}{\sqrt{25+13\left(0\right)+14\left(0\right)+17\left(0\right)+6\left(0\right)}}=\frac{5}{5}=1$