Question

$\text{Evaluate :}\underset{x\to \infty }{\text{lim}}{\left(\frac{2x^2-1}{2x^2+3}\right)}^{4x^2+2}$

• A. $e^{-8}$
• B. $e^{-4}$
• C. $0$
• D. $1$

Answer 1

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

Given :$\underset{x\to \infty }{lim}{\left(\frac{2x^2-1}{2x^2+3}\right)}^{4x^2+2}=l$ (say)

Taking ${\log }_e$ on both sides,

$\ln \left(l\right)=\underset{x\to \infty }{lim}(4x^2+2).\ln \left(\frac{2x^2-1}{2x^2+3}\right)$

$\ln \left(l\right)=\underset{x\to \infty }{lim}(4x^2+2)\ln \left(\frac{2-\frac{1}{x^2}}{2+\frac{3}{x^2}}\right)$

$\ln \left(l\right)=\underset{x\to \infty }{lim}\frac{\ln \left(\frac{2-\frac{1}{x^2}}{2+\frac{3}{x^2}}\right)}{1/(4x^2+2)}$

$\ln \left(l\right)=\underset{x\to \infty }{lim}\frac{\left(\frac{2+\frac{3}{x^2}}{2-\frac{1}{x^2}}\right).[\frac{+\frac{2}{x^3}}{(2+\frac{3}{x^2})}+\frac{(2-\frac{1}{x^2})\left(\frac{6}{x^3}\right)}{{(2+\frac{3}{x^2})}^2}]}{\frac{-8x}{{(4x^2+2)}^2}}$ [L-Hospital Rule]

$\ln \left(l\right)=\underset{x\to \infty }{lim}\frac{\left(\frac{2+\frac{3}{x^2}}{2-\frac{1}{x^2}}\right).\frac{1}{x^3}[\frac{2}{(2+\frac{3}{x^2})}+\frac{(2-\frac{1}{x^2})6}{{(2+\frac{3}{x^2})}^2}]}{\frac{1}{x^3}.\frac{-8}{{(4+\frac{2}{x^4})}^2}}$

$\ln \left(l\right)=\frac{\left(\frac{2}{2}\right).[\left(\frac{2}{2}\right)+\frac{12}{4}]}{\frac{-8}{16}}=-8$

$\Rightarrow \ln \left(l\right)=-8$

$\Rightarrow l=e^{-8}$