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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.

$$? \le \mathop {\lim }\limits_{n \to \infty } {{n - 1} \over {n + 1}} \le \mathop {\lim }\limits_{n \to \infty } {{n + 1} \over {n + 1}} = 1$$

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    $\begingroup$ $$\frac{n-1}{n+1}=1-\frac2{n+1}$$ $\endgroup$ Dec 1, 2013 at 11:09
  • $\begingroup$ The tag (limit-theorems) is intended for questions about limit theorems in probability theory and not for questions about determining limits of sequences or functions, see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) $\endgroup$ Dec 1, 2013 at 14:29

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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.

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  • $\begingroup$ Oops. I got to wake up before doing math $\endgroup$ Dec 1, 2013 at 11:12
  • $\begingroup$ By the way, what happens for the limit $\mathop {\lim (}\limits_{n \to \infty } {{n - 1} \over {n + 1}}{)^{{n^2}}}$ $\endgroup$ Dec 1, 2013 at 11:15
  • $\begingroup$ @DanielGagnon Intuitively the larger power in either the numerator or the denominator dominates the other. Your last ratio goes to infinity. $\endgroup$
    – JohnK
    Dec 1, 2013 at 11:17
  • $\begingroup$ But how can you show it formaly? $\endgroup$ Dec 1, 2013 at 11:28
  • $\begingroup$ I use theorems of kind $\lim \frac ab=\frac{\lim a}{\lim b}$ and $\lim a+b=\lim a+\lim b$ $\endgroup$ Dec 1, 2013 at 11:43
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$$\lim_{n\to\infty}\frac{n-1}{n+1}=\lim_{n\to\infty}\frac{n\left(1-\frac{1}{n}\right)}{n\left(1+\frac{1}{n}\right)}=\lim_{n\to\infty}\frac{1-\frac{1}{n}}{1+\frac{1}{n}}=\frac{1-\lim_{n\to\infty}\frac{1}{n}}{1+\lim_{n\to\infty}\frac{1}{n}}=\frac{1-0}{1+0}=1$$

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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.

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