Question

Помогите решить предел

Answer 1

$\lim_{x \to \infty} (\sqrt{x^2+4x+2} - \sqrt{x^2-2x+4}) = \lim_{x \to \infty} \frac{(\sqrt{x^2+4x+2} - \sqrt{x^2-2x+4})(\sqrt{x^2+4x+2} + \sqrt{x^2-2x+4})}{\sqrt{x^2+4x+2} + \sqrt{x^2-2x+4}} = \lim_{x \to \infty}\frac{x^2+4x+2-(x^2-2x+4)}{x(\sqrt{1 + \frac{4}{x} + \frac{2}{x^2}} + \sqrt{1 - \frac{2}{x} + \frac{4}{x^2}})} =\lim_{x \to \infty} \frac{6x + 6}{{x(\sqrt{1 + \frac{4}{x} + \frac{2}{x^2}} + \sqrt{1 - \frac{2}{x} + \frac{4}{x^2}}})} =$$= \lim_{x \to \infty} \frac{6 + \frac{6}{x}}{\sqrt{1 + \frac{4}{x} + \frac{2}{x^2}} + \sqrt{1 - \frac{2}{x} + \frac{4}{x^2}}} = \frac{6 + 0}{\sqrt{1+0+0} + \sqrt{1-0+0}} = \frac{6}{1 + 1}=3$

Answer 2

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