Question

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

• A. $e^{-8}$
• B. $e^8$
• C. $e^3$
• D. $e^{-3}$

Answer 1

The correct option is A $e^{-8}$

${lim}_{x\to \infty }{\left[\frac{2x^2+3}{2x^2+5}\right]}^{8x^2+3}=l$ ( is of the form ${\left(1\right)}^{\infty }$)

$\Rightarrow l={lim}_{x\to \infty }{\left[\frac{2+\frac{3}{x^2}}{2+\frac{5}{x^2}}\right]}^{8x^2+3}$

Taking ${\log }_e$ on both sides

$\ln l={lim}_{x\to \infty }(8x^2+3)\ln \left[\frac{2+\frac{3}{x^2}}{2+\frac{5}{x^2}}\right]$

$\ln l={lim}_{x\to \infty }\frac{\ln \left[\frac{2+\frac{3}{x^2}}{2+\frac{5}{x^2}}\right]}{\frac{1}{8x^2+3}}$ ( is of the form $\frac{\infty }{0}$)

$\ln l={lim}_{x\to \infty }\frac{\left[\frac{2+\frac{3}{x^2}}{2+\frac{5}{x^2}}\right].[\frac{\frac{3(-2)}{x^3}}{(2+\frac{5}{x^2})}+\frac{(2+\frac{3}{x^2})\left(1\right)\left(\frac{5}{x^3}\right)\left(2\right)}{{(2+\frac{5}{x^2})}^2}]}{\frac{-1\left(8\right)\left(2\right)x}{{(8x^2+3)}^2}}$ (L' Hospital rule)

$\ln l={lim}_{x\to \infty }\frac{\left[\frac{2+\frac{5}{x^2}}{2+\frac{3}{x^2}}\right]\frac{1}{x^3}[\frac{-6}{(2+\frac{5}{x^2})}+\frac{(2+\frac{3}{x^2})\left(10\right)}{{(2+\frac{5}{x^2})}^2}]}{\frac{1}{x^3}\frac{-16}{{(8+\frac{3}{x^2})}^2}}$

$\ln l=\frac{\left(1\right)\left(2\right)}{-16}\times 64$

$\ln l=-8$

$l=e^{-8}$