Question

Найдите предел последовательности

Answer 1

$lim_{n to infty} a_n = lim_{n to infty} (1+frac{1}{n})^{2n}= lim_{n to infty} [(1+frac{1}{n})^n]^2= lim_{n to infty}e^2=e^2$≈$7,389$

$lim_{n to infty} b_n = lim_{n to infty} (1+frac{1}{2n})^{n}= lim_{n to infty} (1+frac{1}{2n})^{2ncdotfrac{1}{2}}= \= lim_{n to infty} [(1+frac{1}{2n})^{2n}]^{frac{1}{2}}=lim_{n to infty}e^{frac{1}{2}}=e^{frac{1}{2}}=sqrt{e}$≈$1,649$

$lim_{n to infty} c_n = lim_{n to infty} (1-frac{1}{n})^{n}= lim_{n to infty} (1+frac{1}{-n})^{-ncdot(-1)}= \= lim_{n to infty} [(1+frac{1}{-n})^{-n}]^{-1}=lim_{n to infty}e^{-1}=e^{-1}=frac{1}{e}$≈$0,368$