Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 146

\[1)\ \lim_{x \rightarrow 0}\frac{3 - x + x^{3}}{2 + 2x - x^{2}} =\]

\[= \frac{3 - 0 + 0}{2 + 0 - 0} = \frac{3}{2};\]

\[2)\ \lim_{x \rightarrow 6}\frac{\sqrt{x - 2} - 2}{x - 6} =\]

\[= \lim_{x \rightarrow 6}\frac{x - 2 - 4}{(x - 6)\left( \sqrt{x - 2} + 2 \right)} =\]

\[= \lim_{x \rightarrow 6}\frac{1}{\sqrt{x - 2} + 2} = \frac{1}{\sqrt{6 - 2} + 2} =\]

\[= \frac{1}{2 + 2} = \frac{1}{4};\]

\[3)\ \lim_{x \rightarrow \infty}\frac{3x^{2} + 4x + 7}{6x^{2} - x + 5} =\]

\[= \lim_{x \rightarrow \infty}\frac{3 + \frac{4}{x} + \frac{7}{x^{2}}}{6 - \frac{1}{x} + \frac{5}{x^{2}}} = \frac{3}{6} = \frac{1}{2};\]

\[4)\ \lim_{x \rightarrow 4}\frac{\sqrt{1 + 2x} - 3}{\sqrt{x} - 2} =\]

\[= \lim_{x \rightarrow 4}\frac{1 + 2x - 9}{\left( \sqrt{x} - 2 \right)\left( \sqrt{1 + 2x} + 3 \right)} =\]

\[= \lim_{x \rightarrow 4}\frac{2\left( \sqrt{x} + 2 \right)}{\sqrt{1 + 2x} + 3} =\]

\[= \frac{2(2 + 2)}{\sqrt{1 + 8} + 3} = \frac{8}{3 + 3} = \frac{8}{6} = \frac{4}{3};\]

\[5)\ \lim_{x \rightarrow + \infty}\frac{\sqrt{3x^{2} + 4x + 7}}{x} =\]

\[= \lim_{x \rightarrow + \infty}\sqrt{3 + \frac{4}{x} + \frac{7}{x^{2}}} = \sqrt{3};\]

\[= \lim_{x \rightarrow + \infty}\frac{(x^{2} + 2x + 3) - (x^{2} - x + 1)}{\sqrt{x^{2} + 2x + 3} + \sqrt{x^{2} - x + 1}} =\]

\[= \lim_{x \rightarrow + \infty}\frac{3x + 2}{\sqrt{x^{2} + 2x + 3} + \sqrt{x^{2} - x + 1}} =\]

\[= \lim_{x \rightarrow + \infty}\frac{3 + \frac{2}{x}}{\sqrt{1 + \frac{2}{x} + \frac{3}{x^{2}}} + \sqrt{1 - \frac{1}{x} + \frac{1}{x^{2}}}} =\]

\[= \frac{3}{1 + 1} = \frac{3}{2}.\]