Question

Решить предел. Предел на фото.

Answer 1

$lim_{x to infty} ln(1 - frac{2}{x} ) ctg frac{1}{x} = lim_{x to infty} frac{ ln(1 - frac{2}{x} ) cos( frac{1}{x} ) }{ sin( frac{1}{x} ) } = \ = lim_{x to infty} cos( frac{1}{x} ) lim_{x to infty} frac{ ln(1 - frac{2}{x} ) }{ sin( frac{1}{x} ) } = \ frac{1}{x} = t; : x to infty Rightarrow t to 0 \ = cos( frac{1}{ infty } ) lim_{t to 0} frac{ ln(1 - 2t) }{t} = lim_{t to 0} frac{( ln(1 - 2t) )'}{(t)'} = \ = lim_{t to 0} frac{ - 2}{1 - 2t} = - 2 times frac{1}{1 - 2 times 0} = - 2$

Ответ: -2.