Задача 69593 Необходимо решить 1.3 и 1.4 задачу, m =...
Условие
Решение
[m] lim_{x → 0 }\frac{tg7x}{sin6x}= lim_{x → 0 }\frac{tg 7x}{7x}\cdot \frac{7x}{sin6x}= lim_{x → 0 }\frac{tg 7x}{7x}\cdot \frac{6x}{sin6x}\cdot \frac{7}{6}=1\cdot 1\cdot \frac{7}{6}=\frac{7}{6}[/m]
1.4
[m] lim_{x → ∞ }(\frac{x+7}{x-3})^{5x-3}=lim_{x → ∞ }(\frac{x+7}{x-3})^{5x}\cdot (\frac{x+7}{x-3})^{-3}=lim_{x → ∞ }(\frac{x+7}{x-3})^{5x}\cdot lim_{x → ∞ }(\frac{x+7}{x-3})^{-3}=lim_{x → ∞ }(\frac{\frac{x+7}{x}}{\frac{x-3}{x}})^{5x} \cdot 1^{-3}=[/m]
[m] =lim_{x → ∞ }\frac{(1+\frac{7}{x})^{5x}}{1-\frac{3}{x})^{5x}} =\frac{lim_{x → ∞ }(1+\frac{7}{x})^{5x}}{lim_{x → ∞ (1-\frac{3}{x})^{5x}}}=\frac{(e^{7})^{5}}{(e^{-3})^{5}}=e^{40}[/m]
