Question

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

Answer 1

${lim}_{x\to \infty }{\left(\frac{1+2x}{1+3x}\right)}^x={lim}_{x\to \infty }{\left(\frac{\frac{1}{x}+2}{\frac{1}{x}+3}\right)}^x={lim}_{x\to \infty }{\left(\frac{2}{3}\right)}^x=0$