Question

${lim}_{x\to \infty }{\left\{\frac{x^2+2x+3}{2x^2+x+5}\right\}}^{\frac{3x-2}{3x+2}}$

Answer 1

$Lety={lim}_{x\to \infty }{\left\{\frac{x^2+2x+3}{2x^2+x+5}\right\}}^{\frac{3x-2}{3x+2}}$
TAking log on both sides
$logy=\frac{3x-2}{3x+2}log\left(\frac{x^2+2x+3}{2x^2+x+5}\right)y=\frac{3x-2}{e^{3x+2}log\left(\frac{x^2+2x+3}{2x^2+x+5}\right)}{\left\{\frac{x^2+62x+3}{2x^2+x+5}\right\}}^{\frac{3x-2}{3x+2}}=e^{\frac{3x-2}{3x+2}log\left(\frac{x^2+2x+3}{2x^2+x+5}\right)}$
taking limit $x\to 0={lim}_{x\to \infty }{\left\{\frac{x^2+2x+3}{2x^2+x+5}\right\}}^{\frac{3x-2}{3x+2}}=e^{{lim}_{x\to }\frac{(3-\frac{2}{x})}{(3+\frac{2}{x})}log\left(\frac{1+\frac{2}{x}+\frac{3}{x^2}}{2+\frac{2}{x}+\frac{5}{x}}\right)}=e^{\left(\frac{3-0}{3+0}\right)log\frac{1}{2}}[\because asx\to \infty ,\frac{1}{x}\to 0]=e^{log\frac{1}{2}}=\frac{1}{2}$