Question

Peco ajuda em limtes em matematica
$\lim_{n \to \infty}( \frac{4x-3}{4x+2}) ^ \frac{x+1}{3}$

Answer 1

$L=\lim\limits_{x \to \infty}( \frac{4x-3}{4x+2}) ^ \frac{x+1}{3} \\ \\ L=\lim\limits_{x \to \infty}( 1-\frac{5}{4x+2}) ^ \frac{x+1}{3} \\ \\ L=\lim\limits_{x \to \infty}( 1-\frac{5}{4x+2}) ^ {-\frac{4x+2}{5}\cdot\frac{x+1}{3}\cdot\frac{5}{4x+2}} \\ \\ L=(1/e)^{\lim\limits_{x\to \infty}\frac{5}{3}\cdot\frac{x+1}{4x+2}}\\ \\ L=(1/e)^{\frac{5}{3}\cdot\frac{1}{4}}\\ \\ \boxed{L=(1/e)^{5/12}}$