Question

lim x стремится к бесконечности ((2x+5)/2x)^4x-1

Answer 1

$\lim_{x \to \infty} ( \frac{2x+5}{2x})^{4x-1} = \lim_{x \to \infty} ( 1+\frac{5}{2x})^{4x-1} = \\ \\ = \lim_{x \to \infty} (( 1+\frac{5}{2x})^{ \frac{2x}{5}})^ { \frac{5}{2x}*(4x-1)} = \lim_{x \to \infty} e^{ \frac{20x-5}{2x}} = \\ \\ = e^{ \lim_{x \to \infty} \frac{20x-5}{2x} }=e^{ \frac{20}{2} }=e^{10}$