Question

Найти пределы не применяя правило Лопиталя
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Answer 1

$\displaystyle 1. \ \lim_{x \to 3} \dfrac{\sqrt{2x + 2} - \sqrt{3x - 1}}{\sqrt{3x+1} - \sqrt{2x + 4}} = \left(\dfrac{0}{0} \right) =$

$\displaystyle = \lim_{x \to 3} \dfrac{(\sqrt{2x + 2} - \sqrt{3x - 1})(\sqrt{2x + 2} + \sqrt{3x - 1})(\sqrt{3x+1} + \sqrt{2x + 4})}{(\sqrt{3x+1} - \sqrt{2x + 4})(\sqrt{3x+1} + \sqrt{2x + 4})(\sqrt{2x + 2} + \sqrt{3x - 1})} =$

$\displaystyle = \lim_{x \to 3} \dfrac{((\sqrt{2x + 2})^{2} - (\sqrt{3x - 1})^{2})(\sqrt{3x+1} + \sqrt{2x + 4})}{((\sqrt{3x+1})^{2} - (\sqrt{2x + 4})^{2})(\sqrt{2x + 2} + \sqrt{3x - 1})} =$

$\displaystyle = \lim_{x \to 3} \dfrac{(2x + 2 - (3x - 1))(\sqrt{3x+1} + \sqrt{2x + 4})}{(3x + 1 - (2x + 4))(\sqrt{2x + 2} + \sqrt{3x - 1})} =$

$\displaystyle = \lim_{x \to 3} \dfrac{(3 - x)(\sqrt{3x+1} + \sqrt{2x + 4})}{(x-3)(\sqrt{2x + 2} + \sqrt{3x - 1})} = \lim_{x \to 3} -\dfrac{\sqrt{3x+1} + \sqrt{2x + 4}}{\sqrt{2x + 2} + \sqrt{3x - 1}} =$

$=-\dfrac{\sqrt{3 \cdot 3+1} + \sqrt{2\cdot 3 + 4}}{\sqrt{2\cdot 3 + 2} + \sqrt{3\cdot 3 - 1}} = -\dfrac{\sqrt{10} + \sqrt{10}}{\sqrt{8} + \sqrt{8}} =-\dfrac{\sqrt{5}}{2}$

$2. \ \displaystyle \lim_{x \to 0} \dfrac{x^{3} - 5x^{2}}{1 - \cos 3x} = \left(\dfrac{0}{0} \right) = \lim_{x \to 0} \dfrac{x^{3} - 5x^{2}}{1 - \cos \left(2 \cdot \dfrac{3x}{2} \right)} =$

$= \displaystyle \lim_{x \to 0} \dfrac{x^{3} - 5x^{2}}{1 - \left(1 - 2\sin^{2}\dfrac{3x}{2} \right)} = \lim_{x \to 0} \dfrac{x^{3} - 5x^{2}}{2\sin^{2}\dfrac{3x}{2} } = \left|\begin{array}{ccc}\sin \dfrac{3x}{2} \sim \dfrac{3x}{2} \\x \to 0 \\\end{array}\right| =$

$= \displaystyle \lim_{x \to 0} \dfrac{x^{3} - 5x^{2}}{2 \cdot \left(\dfrac{3x}{2} \right)^{2}} = \lim_{x \to 0} \dfrac{2x^{2}(x - 5)}{9x^{2}} = \lim_{x \to 0} \dfrac{2x - 10}{9} = -\dfrac{10}{9}$