Prove $\mathop {\lim }\limits_{n \to \infty } {{n - 1} \over {n + 1}} = 1$ I tried to use the squeezing principle, but couldn't find a proper expression for the left inequality. Maybe there's another way, it seems simple though I couldn't figure it out.
$$? \le \mathop {\lim }\limits_{n \to \infty } {{n - 1} \over {n + 1}} \le \mathop {\lim }\limits_{n \to \infty } {{n + 1} \over {n + 1}} = 1$$
 A: Why not to show it directly:
$$\mathop {\lim }\limits_{n \to \infty } {{n - 1} \over {n + 1}}=\mathop {\lim }\limits_{n\to \infty } {{1 - \frac1n} \over {1 + \frac1n}}=1$$
This inferred from theorems for limit of quotient and limit of difference/sum.
EDIT:You can use the limit definition:
You need to show that $$(\forall\epsilon>0)(\exists n_0>0)(\forall n>n_0):\left|{{n - 1} \over {n + 1}}-1 \right|<\epsilon$$
$$(\forall\epsilon>0)(\exists n_0>0)(\forall n>n_0):\left|{{n - 1 -(n+1)} \over {n + 1}} \right|<\epsilon$$
$$(\forall\epsilon>0)(\exists n_0>0)(\forall n>n_0):{|{-2}| \over |{n + 1}|} <\epsilon$$
$$(\forall\epsilon>0)(\exists n_0>0)(\forall n>n_0):{2 \over {n + 1}} <\epsilon$$
$$(\forall\epsilon>0)(\exists n_0>0)(\forall n>n_0):{2 \over \epsilon} <{n+1}$$
Take $n_0=\lceil 2/e \rceil$ and the ineqality holds.
A: $$\lim_{n\to\infty}\frac{n-1}{n+1}=\lim_{n\to\infty}\frac{n\left(1-\frac{1}{n}\right)}{n\left(1+\frac{1}{n}\right)}=\lim_{n\to\infty}\frac{1-\frac{1}{n}}{1+\frac{1}{n}}=\frac{1-\lim_{n\to\infty}\frac{1}{n}}{1+\lim_{n\to\infty}\frac{1}{n}}=\frac{1-0}{1+0}=1$$
A: How about L'Hopital's rule? After all when $n$ goes to infinity you have an indefinite ratio $\frac{\infty}{\infty}$ since the constants are negligent.
