Question

Lim x -> бесконечность ((x^2-1)/(x^2+1))^((x-1)/(x+1)) через формулу lim f (x)^g(x)=e^lim g(x) ln f(x)

Answer 1

$\lim_{x \to \infty }( \frac{ {x}^{2} - 1}{ {x}^{2} + 1} ) {}^{ \frac{x - 1}{x + 1} } =\lim_{x \to \infty } {e}^{ ln(( \frac{ {x}^{2} - 1}{ {x}^{2} + 1} ) {}^{ \frac{x - 1}{x + 1} } ) } = {e}^{ \lim_{x \to \infty } \frac{x - 1}{x + 1} ln( \frac{ {x}^{2} - 1}{ {x}^{2} + 1} ) } = {e}^{ \lim_{x \to \infty }( \frac{x - 1}{x + 1}) \lim_{x \to \infty }( ln( \frac{ {x}^{2} - 1 }{ {x}^{2} + 1} ) )}$

Вычислим пределы по-отдельности:

$\lim_{x \to \infty } \frac{x - 1}{x + 1} = \lim_{x \to \infty } \frac{x(1 - \frac{1}{x} )}{x(1 + \frac{1}{x}) } = \lim_{x \to \infty } = \frac{1 - \frac{1}{x} }{1 + \frac{1}{x} } = \frac{1 - \frac{1}{ \infty } }{1 + \frac{1}{ \infty } } = 1$

$\lim_{x \to \infty } ln( \frac{ {x}^{2} - 1 }{ {x}^{2} + 1 } ) = ln( \lim_{x \to \infty } \frac{ {x}^{2} - 1}{ {x}^{2} + 1} ) = ln( \lim_{x \to \infty } \frac{ {x}^{2}(1 - \frac{1}{ {x}^{2} } ) }{ {x}^{2}(1 + \frac{1}{ {x}^{2} } )} ) = ln( \lim_{x \to \infty } \frac{1 - \frac{1}{ {x}^{2} } }{1 + \frac{1}{ {x}^{2} } } ) = ln( \frac{1 - \frac{1}{ \infty } }{1 + \frac{1}{ \infty } } ) = ln(1) = 0$

Собираем:

$\lim_{x \to \infty }( \frac{ {x}^{2} - 1}{ {x}^{2} + 1} ) {}^{ \frac{x - 1}{x + 1} } = {e}^{1 \times 0} = {e}^{0} = 1$

Ответ: 1