Предмет: Алгебра, автор: probaproba99

Вычислить пределы функции. 98Б
Задание на фотографии. Спасибо! :3

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Автор ответа: Alexandr130398

0

$1)lim_{x to 3} frac{ x^{2} -9}{ x^{2} -8x+15} =lim_{x to 3} frac{ (x-3)(x+3)}{ (x-3)(x-5)}=lim_{x to 3} frac{(x+3)}{(x-5)}= \ frac{3+3}{3-5} = frac{6}{-2} =-3 \2) lim_{x to 2} frac{x+5}{3x-6}= lim_{x to 2} frac{2+5}{3*2-6} = lim_{x to 2} frac{7}{0} =oo$
oo-это бесконечность

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