Задача 72958 вычислить приделы функций...
Условие
Решение
1) [m]\lim \limits_{x \to 2} \frac{7x - 4 + x^2}{3x + 2x^2 - 1} = \frac{7 \cdot 2 - 4 + 2^2}{3 \cdot 2 + 2 \cdot 2^2 - 1} = \frac{14 - 4 + 4}{6 + 8 - 1} = \frac{14}{13} [/m]
2) [m]\lim \limits_{x \to 1} \frac{7x^2 - 8x + 1}{x^2 - 1} = \lim \limits_{x \to 1} \frac{(x-1)(7x-1)}{(x-1)(x+1)} =\lim \limits_{x \to 1} \frac{7x-1}{x+1} = \frac{7 \cdot 1-1}{1+1} = \frac{6}{2} = 3[/m]
3) [m]\lim \limits_{x \to \infty} \frac{4x^2 - 3x + 2}{x^2 + 2x + 5} = \lim \limits_{x \to \infty} \frac{4 - 3/x + 2/x^2}{1 + 2/x + 5/x^2} = \frac{4 - 0 + 0}{1 + 0 + 0} = 4[/m]
4) [m]\lim \limits_{x \to 0} \frac{sin(3x)}{sin(8x)} = \lim \limits_{x \to 0} \frac{sin(3x)}{3x} \cdot \frac{8x}{sin(8x)} \cdot \frac{3x}{8x} = 1 \cdot 1 \cdot \frac{3}{8} = \frac{3}{8}[/m]
5) [m]\lim \limits_{x \to \infty} (\frac{4x + 1}{4x - 3})^{x} = \lim \limits_{x \to \infty} (\frac{4x - 3 + 4}{4x - 3})^{x} =\lim \limits_{x \to \infty} (1 + \frac{4}{4x - 3})^{x} =\lim \limits_{x \to \infty} (1 + \frac{4}{4x - 3})^{(4x-3+3)/4} = [/m]
[m]= \lim \limits_{x \to \infty} (1 + \frac{4}{4x - 3})^{(4x-3)/4} \cdot (1 + \frac{4}{4x - 3})^{3/4} = e^{4/4} \cdot (1 + 0)^{3/4} = e[/m]
6) [m]\lim \limits_{x \to 4} \frac{\sqrt{x+5}-3}{x^2-16} = \lim \limits_{x \to 4} \frac{(\sqrt{x+5}-3)(\sqrt{x+5}+3)}{(x-4)(x+4)(\sqrt{x+5}+3)} = \lim \limits_{x \to 4} \frac{x+5-9}{(x-4)(x+4)(\sqrt{x+5}+3)} =[/m]
[m]= \lim \limits_{x \to 4} \frac{1}{(x+4)(\sqrt{x+5}+3)} = \frac{1}{(4+4)(\sqrt{4+5}+3)} =\frac{1}{8(3+3)} =\frac{1}{48}[/m]