Граница на функция

[Гост]

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[Davids]

[tex]A = \frac{1}{x+3} - \frac{27}{x^3 + 3^3} =[/tex]
$= \frac{1}{x+3} - \frac{27}{(x+3)(x^2-3x+9)} =$
$= \frac{x^2-3x-18}{(x+3)(x^2-3x+9)} = \frac{(x+3)(x-6)}{(x+3)(x^2-3x+9)} = \frac{x-6}{x^2-3x+9}$
$\Rightarrow \lim_{x \to -3}A = \frac{-9}{27} = -\frac{1}{3}$