Question

Решить интеграл ( хотя бы 1)

Answer 1

$\lim_{n \to \infty} \frac{5n ^{6}+1-n }{n ^{2} -n ^{3} +1}=\frac{\infty}{\infty}=\lim_{n \to \infty} \frac{n^6(5+\frac{1}{n^6}^{\rightarrow0}-\frac{1}{n^5}^{\rightarrow0}) }{n^6(\frac{1}{n^4}^{\rightarrow0} -\frac{1}{n^3}^{\rightarrow0} +\frac{1}{n^6}^{\rightarrow0})}=\frac{5}{0}=\infty$

$\lim_{x \to 1} \frac{2x ^{2}-14x+12 }{ x^{2} -6x+5}=\frac{0}{0}=\lim_{x \to 1}\frac{2(x^2-7x+6)}{(x^2-6x+5)}=2lim_{x \to 1}\frac{(x-6)(x-1)}{(x-5)(x-1)}=\\=2lim_{x \to 1}\frac{(x-6)}{(x-5)}=2*\frac{5}{4}=2,5$

$\lim_{x \to 5} \frac{ \sqrt{x-1}-2 }{x-5} =\frac{0}{0}=\lim_{x \to 5} \frac{ \sqrt{x-1}-2 }{x-5}*\frac{\sqrt{x-1}+2}{\sqrt{x-1}+2}=\\\lim_{x \to 5}\frac{x-5}{(x-5)*(\sqrt{x-1}+2)}}=\lim_{x \to 5}\frac{1}{\sqrt{x-1}+2}=\frac{1}{4}=0,25$

$\displaystyle \lim_{n \to \infty} (\frac{n}{1+n})^n=(\frac{\infty}{\infty})^\infty= \lim_{n \to \infty} (\frac{1+n-1}{1+n})^n=\\=\lim_{n \to \infty} (1+\frac{1}{-(1+n)})^n=\lim_{n \to \infty} [(1+\frac{1}{-(1+n)})^{-(1+n)}]^{-\frac{n}{1+n}}=\\=e^{-\lim_{n \to \infty}(\frac{n}{n+1})}=e^{-\lim_{n \to \infty}(\frac{n}{n(1+\frac{1}{n}^{\to0})})}=e^{-1}=\frac{1}{e}$