Question

Va rog ajutati-ma la exercitiu

Answer 1

Răspuns:

$\lim_{x\to\infty} \left(\frac{x+1}{x+3} \right)^{2x+1}=\lim_{x\to\infty} \left(1+\frac{-2}{x+3} \right)^{2x+1}=\lim_{x\to\infty} \left(\left(1+\frac{-2}{x+3}\right)^{\frac{x+3}{-2}}\right)^{\frac{-2(2x+1)}{x+3}}=e^{-4}$.