Question

$\underset{x\to \infty }{lim}(\frac{5x^2+x+3}{x^2+x+3}{)}^{\frac{1}{x}}$
(2002)

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

Answer 1

The correct option is A

$e^4$

$\underset{x\to \infty }{lim}\left\{\begin{array}{c}\frac{5x^2+x+3}{x^2+x+3}\end{array}\right\}=\underset{x\to \infty }{lim}{\left\{\begin{array}{c}\frac{5+1/x+3/x^2}{1+1/x+3/x^2}\end{array}\right\}}^x$

$=\left[1^{\infty }Form\right]$

Hence, limit $=e^{\underset{x\to \infty }{lim}}\left\{\begin{array}{c}\frac{5x^2+x+3}{x^2+x+3}-1\end{array}\right\}x$

$=e^{\underset{x\to \infty }{lim}}\left\{\begin{array}{c}\frac{4x^2}{x^2+x+3}\end{array}\right\}$
[Using L’ Hospital Rule]
$=e^{\underset{x\to \infty }{lim}}\left\{\begin{array}{c}\frac{8x}{2x+1}\end{array}\right\}$
$=e^{\underset{x\to \infty }{lim}}\left\{\begin{array}{c}\frac{8}{2}\end{array}\right\}$
$=e^4$