Question

Вычислить пределы, не используя правило Лопиталя

Answer 1

Ответ:

Предел:

$\boldsymbol{\boxed{ \lim_{x \to 7} \frac{2x^{2} - 11x - 21}{x^{2} - 9x + 14} = 3,4}}$

Пошаговое объяснение:

$\displaystyle \lim_{x \to 7} \frac{2x^{2} - 11x - 21}{x^{2} - 9x + 14} =$

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а)

$2x^{2} - 11x - 21 = 0$

$D = 121 - 4 \cdot 2 \cdot (-21) = 121 + 168 = 289 = 17^{2}$

$x_{1} = \dfrac{11 + 17}{2 \cdot 2} = \dfrac{28}{4} = 7$

$x_{2} = \dfrac{11 - 17}{2 \cdot 2} = \dfrac{-6}{2 \cdot 2} = -\dfrac{3}{2} = -1,5$

$2x^{2} - 11x - 21 = 2(x - 7)(x +1,5)$

б)

$x^{2} - 9x + 14 = 0$

$D = 81 - 4 \cdot 1 \cdot 14 = 81 - 56 = 25 = 5^{2}$

$x_{1} = \dfrac{9 + 5}{2} = \dfrac{14}{2} = 7$

$x_{2} = \dfrac{9 - 5}{2} = \dfrac{4}{2} = 2$

$x^{2} - 9x + 14 = (x - 7)(x - 2)$

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$\displaystyle = \lim_{x \to 7} \frac{2(x - 7)(x +1,5)}{ (x - 7)(x - 2)} = \lim_{x \to 7} \frac{2x + 3}{x - 2} = \frac{2 \cdot 7 + 3}{7 - 2} = \frac{14 + 3}{5} = \frac{17}{5} = 3,4$