Question

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Answer 1

$\displaystyle \lim_{x \to 3} \frac{x^3-27}{x-3}=\frac{0}{0}= \lim_{x \to 3} \frac{(x-3)(x^2+3x+9)}{x-3}= \lim_{x \to 3} x^2+3x+9=\\=27$

Answer 2

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