Задача 78160 нужна помощь \0_0/...
Условие
Решение
1) [m]\lim \limits_{x \to \infty} (5 + \frac{2}{x} - \frac{3}{x^2}) = 5 + \frac{2}{\infty} - \frac{3}{\infty} = 5 + 0 - 0 = 5[/m]
2) [m]\lim \limits_{x \to \infty} (\frac{3x^3-x}{7+2x^3}) = \lim \limits_{x \to \infty} (\frac{3-1/x^2}{7/x^3+2}) =\lim \limits_{x \to \infty} (\frac{3-0}{0+2}) =\frac{3}{2}[/m]
3) [m]\lim \limits_{x \to \infty} (\frac{5x^3-3x}{2+x^2}) = \lim \limits_{x \to \infty} (\frac{5-3/x^2}{2/x^3+1/x}) = (\frac{5-0}{0+0}) = \infty[/m]
4) [m]\lim \limits_{x \to \infty} (\frac{x^2+8x-1}{x^5+7x^3+11}) = \lim \limits_{x \to \infty} (\frac{1/x^3+8/x^4-1/x^5}{1+7/x^2+11/x^5}) = (\frac{0+0-0}{1+0+0}) = 0[/m]
5) [m]\lim \limits_{x \to \infty} (\frac{5x^3-x}{7+x^3}) = \lim \limits_{x \to \infty} (\frac{5-1/x^2}{7/x^3+1}) = \frac{5-0}{0+1} = 5[/m]
2 номер
1) [m]\lim \limits_{x \to 0} \frac{\sin 2x}{x} = \lim \limits_{x \to 0} \frac{\sin 2x}{2x} \cdot 2 = 1 \cdot 2 = 2[/m]
2) [m]\lim \limits_{x \to 0} \frac{\sin 5x}{4x} = \lim \limits_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{5x}{4x} = 1 \cdot \frac{5}{4} = \frac{5}{4}[/m]
3) [m]\lim \limits_{x \to 0} \frac{\sin (5x/2)}{x} = \lim \limits_{x \to 0} \frac{\sin (5x/2)}{5x/2} \cdot \frac{5}{2} = 1 \cdot \frac{5}{2} = \frac{5}{2}[/m]
4) [m]\lim \limits_{x \to \infty} (1 + \frac{1}{x})^{5x} = \lim \limits_{x \to \infty} ((1 + \frac{1}{x})^{x})^5 = e^5[/m]
5) [m]\lim \limits_{x \to \infty} (1 + \frac{3}{x})^{x/3 \cdot 3} = \lim \limits_{x \to \infty} ((1 + \frac{3}{x})^{x/3})^3 = e^3[/m]