Задача 80675 Найти пределы...
Условие
Решение
б) [m]\lim \limits_{x \to 5} \frac{\sqrt{1+3x} - \sqrt{2x+6}}{x^2-5x} = \lim \limits_{x \to 5} \frac{(\sqrt{1+3x} - \sqrt{2x+6})(\sqrt{1+3x} + \sqrt{2x+6})}{(x^2-5x)(\sqrt{1+3x} + \sqrt{2x+6})} =[/m]
[m]= \lim \limits_{x \to 5} \frac{1 + 3x - 2x - 6}{x(x-5)(\sqrt{1+3x} + \sqrt{2x+6})} = \lim \limits_{x \to 5} \frac{x - 5}{x(x-5)(\sqrt{1+3x} + \sqrt{2x+6})} = \lim \limits_{x \to 5} \frac{1}{x(\sqrt{1+3x} + \sqrt{2x+6})}= [/m]
[m]= \frac{1}{5(\sqrt{1+15} + \sqrt{10+6})} = \frac{1}{5(4 + 4)} = \frac{1}{40}[/m]
в) [m]\lim \limits_{x \to 0} \frac{1- \cos 4x}{2x \cdot tg\ 2x} = \lim \limits_{x \to 0} \frac{1- (1 - 2 \sin^2 2x)}{2x \cdot tg\ 2x} = \lim \limits_{x \to 0} \frac{2 \sin^2 2x \cdot \cos 2x}{2x \cdot \sin 2x}= [/m]
[m]= \lim \limits_{x \to 0} \frac{2\sin 2x \cdot \cos 2x}{2x} = \lim \limits_{x \to 0} 2\cos 2x \cdot \lim \limits_{x \to 0} \frac{\sin 2x}{2x} = \lim \limits_{x \to 0} 2\cos 2x \cdot 1 = 2 \cos 0 = 2[/m]
г) [m]\lim \limits_{x \to \infty} \Big ( \frac{2x-5}{2x+1} \Big )^{x+1} = \lim \limits_{x \to \infty} \Big ( \frac{2x+1-6}{2x+1} \Big )^{(2x+1) \cdot \frac{x+1}{2x+1}} = \lim \limits_{x \to \infty} \Big (1 + \frac{-6}{2x+1} \Big )^{(2x+1) \cdot \frac{x+1}{2x+1}} =[/m]
[m]= \lim \limits_{x \to \infty} e^{-6 \cdot \frac{x+1}{2x+1}} = e^{-6 \cdot \lim \limits_{x \to \infty} \frac{x+1}{2x+1}} = e^{-6 \cdot \frac{1}{2}} = e^{-3} = \frac{1}{e^3}[/m]
д) [m]\lim \limits_{x \to 1} \frac{11x^2 + 9x - 20}{x^5-x^4-x^3+2x^2-x} = \lim \limits_{x \to 1} \frac{(x - 1)(11x + 20)}{(x - 1)(x^4 - x^2 + x)} = \lim \limits_{x \to 1} \frac{11x + 20}{x^4 - x^2 + x} = \frac{11 + 20}{1^4 - 1^2 + 1} = \frac{31}{1} = 31[/m]
е) [m]\lim \limits_{x \to \infty} \frac{1- x^8}{1 - x^7} = \lim \limits_{x \to \infty} \frac{x^8 - 1}{x^7 - 1} = \lim \limits_{x \to \infty} \frac{1 - 1/x^8}{1/x - 11/x^8} = \frac{1 - 0}{0 - 0} = \infty[/m]