lim _(xarrow infty )((x+1)/(x-1))^4x+1

Для решения данного предела, мы можем использовать правило Лопиталя и свойство степеней.<br /><br />Сначала, мы можем переписать выражение в виде $\left(\frac{1+1/x}{1-1/x}\right)^{4x+1}$.<br /><br />Затем, мы можем применить правило Лопиталя, чтобы найти предел:<br /><br />$\lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x}\right)^{4x+1} = \lim_{x\to\infty}\left(\frac{1+1/x}{1-1/x