Question

Решите, пожалуйста прошу вас

Answer 1

$displaystyle lim_{x to 7} frac{5x-35}{x^2-8x+7}=lim_{x to 7} frac{5(x-7)}{(x-4)^2-9} =lim_{x to 7} frac{5(x-7)}{(x-4-3)(x-4+3)} =\ \ \ =lim_{x to 7} frac{5(x-7)}{(x-7)(x-1)} =lim_{x to 7} frac{5}{x-1}= frac{5}{7-1}= frac{5}{6}$