Граница на функция

[veskorachev]

Здравейте,
Моля да ми помогнете със следните задачи:

1) lim(n-1/n+1)^n, при n->∞ Отговор: 1/е^2

2) lim √n(√n+1-√n) n+1 цялото е под корена, n->∞

3) lim x-2/2-√x+2 x+2 цялото е под корена, x->2

[kmitov]

[tex]\lim_{n \to \infty}\left(\frac{n-1}{n+1}\right)^n=\lim_{n \to \infty}\left(\frac{n+1-2}{n+1}\right)^n=\lim_{n \to \infty}\left(1-\frac{2}{n+1}\right)^n=\lim_{n \to \infty}\frac{\left(1-\frac{2}{n+1}\right)^{n+1}}{1-\frac{2}{n+1}}[/tex]
[tex]\lim_{n \to \infty}\frac{\left[\left(1-\frac{2}{n+1}\right)^{-(n+1)/2}\right]^{-2}}{1-\frac{2}{n+1}}=\frac{e^{-2}}{1}=e^{-2},[/tex]
като се използва основната граница [tex]\lim_{x \to \infty}(1+1/x)^x=e.[/tex]

На другите задачи не се разбитат условията.

[Добромир Глухаров]

[tex]2.)\ \lim_{n\to\infty}\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)=[\infty.0]=\lim_{n\to\infty}\frac{\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\lim_{n\to\infty}\frac{\sqrt{n}\left(n+1-n\right)}{\sqrt{n+1}+\sqrt{n}}=\\=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{{\lim_{n\to\infty}}\left(\sqrt{\frac{n+1}{n}}+\sqrt{\frac{n}{n}}\right)}=\frac{1}{{\lim_{n\to\infty}}\left(\sqrt{1+\frac{1}{n}}+1\right)}=\\=\frac{1}{\sqrt{1+{\lim_{n\to\infty}}\frac{1}{n}}+1}=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2}[/tex]

[tex]3.)\ \lim_{x\to 2}\frac{x-2}{2-\sqrt{x+2}}=\lim_{x\to 2}\frac{\left(x-2\right)\left(2+\sqrt{x+2}\right)}{\left(2-\sqrt{x+2}\right)\left(2+\sqrt{x+2}\right)}=\lim_{x\to 2}\frac{\left(x-2\right)\left(2+\sqrt{x+2}\right)}{4-x-2}=\\=\lim_{x\to 2}\frac{\left(\cancel{x-2}\right)\left(2+\sqrt{x+2}\right)}{\cancel{2-x}}=-\lim_{x\to 2}\left(2+\sqrt{x+2}\right)=-\left(2+\sqrt{2+2}\right)=-2-\sqrt{4}=-4[/tex]