Question

$\underset{x\to \infty }{lim}{\left(\frac{x^2+5x+3}{x^2+x+3}\right)}^{1/x}$ is

• A. $e^4$
• B. $e^2$
• C. $e^3$
• D. $1$

Answer 1

The correct option is D $1$

$\underset{x\to \infty }{lim}{\left(\frac{x^2+5x+3}{x^2+x+3}\right)}^{\frac{1}{x}}$
Taking $x^2$ common from numerator and denominator,
$={lim}_{x\to \infty }{\left(\frac{1+\frac{5}{x}+\frac{3}{x^2}}{1+\frac{1}{x}+\frac{3}{x^2}}\right)}^{1/x}$
$=1^0=1$
Hence, option 'D' is correct.