Question

решить интеграл ( хотя бы 1)
$lim_{n to infty} frac{5n ^{6}+1-n }{n ^{2} -n ^{3} +1} lim_{n to 1} frac{2x ^{2}-14+12 }{ x^{2} -6x+5} lim_{n to 5} frac{ sqrt{x-1-2} }{x-5} lim_{n to infty} ( frac{n}{1+n} ) ^{n$

Answer 1

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

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

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

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