Question

$\underset{x\to \infty }{lim}{\left[\frac{2x^2+3}{2x^2+5}\right]}^{8x^2+3}$ is equal to

• A. $-8$
• B. $1$
• C. $e^8$
• D. $e^{-8}$

Answer 1

Answer: (D)

$e^{-8}$

$\underset{x\to \infty }{lim}{\left(\frac{2x^2+3}{2x^2+5}\right)}^{8x^2+3}=e^{\left(\underset{x\to \infty }{lim}\left((8x^2+3)\log \left(\frac{2x^2+3}{2x^2+5}\right)\right)\right)}$

$=e^{\left(\underset{x\to \infty }{lim}\left(\frac{\log \left(\frac{2x^2+3}{2x^2+5}\right)}{\frac{1}{(8x^2+3)}}\right)\right)}$

Applying L'Hospital's rule

$=exp\left(\underset{x\to \infty }{lim}-\frac{{(8x^2+3)}^2}{2(2x^2+3)(2x^2+5)}\right)$

Again applying L'Hospital's rule

$=e^{\left(\underset{x\to \infty }{lim}\frac{-3-8x^2}{x^2+2}\right)}=e^{\left(\underset{x\to \infty }{lim}\frac{-\frac{3}{x^2}-8}{1+\frac{2}{x^2}}\right)}=e^{-8}$