Question

Вычислить предел функции ​

Answer 1

$-\frac{1}{4\sqrt{2}}$

Пошаговое объяснение:

$\lim_{x \to -5} \frac{\sqrt{3x+17}-\sqrt{2x+12}}{x^{2}+8x+15}=\lim_{x \to -5} \frac{(\sqrt{3x+17}-\sqrt{2x+12}) \cdot (\sqrt{3x+17}+\sqrt{2x+12})}{(x+3) \cdot (x+5) \cdot (\sqrt{3x+17}+\sqrt{2x+12})}=$

$=\lim_{x \to -5} \frac{(\sqrt{3x+17})^{2}-(\sqrt{2x+12})^{2}}{(x+3) \cdot (x+5) \cdot (\sqrt{3x+17}+\sqrt{2x+12})}=\lim_{x \to -5} \frac{3x+17-(2x+12)}{(x+3) \cdot (x+5) \cdot (\sqrt{3x+17}+\sqrt{2x+12})}=$

$=\lim_{x \to -5} \frac{3x+17-2x-12}{(x+3) \cdot (x+5) \cdot (\sqrt{3x+17}+\sqrt{2x+12})}=\lim_{x \to -5} \frac{x+5}{(x+3) \cdot (x+5) \cdot (\sqrt{3x+17}+\sqrt{2x+12})}=$

$=\lim_{x \to -5} \frac{1}{(x+3) \cdot (\sqrt{3x+17}+\sqrt{2x+12})}=\frac{1}{(-5+3) \cdot (\sqrt{3 \cdot (-5)+17}+\sqrt{2 \cdot (-5)+12}}=$

$=\frac{1}{-2 \cdot (\sqrt{2}+\sqrt{2})}=\frac{1}{-2 \cdot 2\sqrt{2}}=\frac{1}{-4\sqrt{2}}=-\frac{1}{4\sqrt{2}};$