от martin123456 » 10 Яну 2013, 11:24
1.
Знаем, че [tex]\lim_{x\to \infty}(1+\frac{1}{x})^x = e[/tex].
Тогава [tex]\lim_{x\to \infty}(1+\frac{k}{x})^x=\lim_{x\to \infty}{[(1+\frac{1}{\frac{x}{k}})^{\frac{x}{k}}]^k}=e^k[/tex].
[tex]\frac{x-2}{x+5}=\frac{x+5-7}{x+5}=1+\frac{-7}{x+5}[/tex]
[tex]\lim_{x\to \infty}{(\frac{x-2}{x+5})^x}=\lim_{x\to\infty}{(1+\frac{-7}{x+5})^x}=\frac{\lim_{x\to\infty}{(1+\frac{-7}{x+5})^{x+5}}}{\lim_{x\to\infty}{(1+\frac{-7}{x+5})^5}}=\frac{e^{-7}}{1}=e^{-7}[/tex]
2.
[tex]y'=e^{x^2+2x+29}(x^2+2x+29)'+\frac{1}{\sin^2{7x}}(7x)'=e^{x^2+2x+29}(2x+2)+\frac{7}{\sin^2{7x}}[/tex]