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What is the easiest way to evaluate this limit? $\displaystyle{\lim_{n \to\infty} \left(n-1 \over 2n+2\right)^n}$ $$ \text{Is this possible ?$\,$:}\quad \lim_{n \to\infty}\left(n/n - 1/n \over 2n/n + 2/n\right)^n = \lim_{n \to\infty}\left(1 - 1/n \over 2 + 2/n\right)^{n} = \lim_{n \to\infty} \left(1 \over 2\right)^{n} = 0 $$

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  • $\begingroup$ What if it was $\left(\frac{n-1}{n+2}\right)^n$? $\endgroup$ – Siméon Nov 25 '13 at 15:18
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    $\begingroup$ The last line is wrong: you have $\lim \ (1-{1 \over n})^n=\exp(-1)$. Also, in all your limits, the variable is $n$, not $x$. Apart from that, your result is trivially true, since ${{n-1} \over {2n+2}}<1/2$. $\endgroup$ – Jean-Claude Arbaut Nov 25 '13 at 15:19
  • $\begingroup$ @LukaToni You have the limit of $x \to \infty$, but the expression is in terms of $n$ so the answer would simply be $({\frac{n - 1}{2n + 2}})^n$. $\endgroup$ – Zhoe Nov 25 '13 at 15:24
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    $\begingroup$ "Solve" is the wrong word. "Evaluate" would be appropriate. One solves problems; one solves equations; one evaluates expressions. $\endgroup$ – Michael Hardy Nov 25 '13 at 16:14
  • $\begingroup$ The easiest way is to ask Mathematica perhaps. These answers are all next to the easiest at best. $:)$ $\endgroup$ – Ali Nov 25 '13 at 19:37
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For simple limits like this, the easiest way is often to sandwich the sequence between simpler sequences for which the limit is known.

Notice that $n-1 \leq n$ and $2n+2 \geq 2n$ entail $$ \frac{n-1}{2n+2} \leq \frac{1}{2} $$ so that for $n \geq 1$ one has $$0 \leq \left(\frac{n-1}{2n+2}\right)^n \leq \left(\frac{1}{2}\right)^n \xrightarrow[n\to\infty]{} 0.$$ Conclude with the squeeze theorem.

Remark. This method gives more than just the limit: you get a good estimate of how quickly the sequence converges. For example, it is now easy to give a $n$ such that $\left(\frac{n-1}{2n+2}\right)^n \leq 10^{-100}$.

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The limit can be written as $$\frac{\left(\lim_{n\to\infty}\left(1-\frac1n\right)^{-n}\right)^{-1}}{\lim_{n\to\infty} 2^n\cdot \lim_{n\to\infty}\left(1+\frac1n\right)^n}$$

Now use $$\lim_{m\to\infty}\left(1+\frac1m\right)^m=e$$

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  • $\begingroup$ Is this incorrect, another joke, or a legitimate alternative solution? $\endgroup$ – kaine Nov 25 '13 at 15:31
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    $\begingroup$ @kaine, why? It becomes $$\frac{e^{-1}}{\infty\cdot e}=0$$ $\endgroup$ – lab bhattacharjee Nov 25 '13 at 15:32
  • $\begingroup$ Ok, that is correct.. I didn't follow the solution at first but now it makes sense. $\endgroup$ – kaine Nov 25 '13 at 15:35
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$0\le\left(\dfrac{n-1}{2n+2}\right)^n=\dfrac{1}{2^n}\times\dfrac{\left(1-\dfrac{1}{n}\right)^n}{\left(1+\dfrac{1}{n}\right)^n}\le\dfrac{1}{2^n}\times\dfrac{1}{\left(1+\dfrac{1}{n}\right)^n}\to0$ since $\left(1+\dfrac{1}{n}\right)^n\to e$

So by squeezing $\displaystyle{\lim_{n \to\infty} \left(n-1 \over 2n+2\right)^n}=0.$

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \color{#0000ff}{\large\lim_{n \to\infty} \left(n - 1 \over 2n + 2\right)^{n}} = \lim_{n \to\infty}\bracks{{1 \over 2^{n}}\, {\pars{1 - 1/n}^{n} \over \pars{1 + 1/n}^{n}}} = \lim_{n \to\infty}\pars{{1 \over 2^{n}}\,{\expo{-1} \over e}} = \color{#0000ff}{\large 0} $$

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$$\frac{n-1}{2n+2} = \frac{1}{2}\frac{n+1}{n+1}-\frac{1}{2}\frac{2}{n+1}$$ $$\lim _{n \rightarrow \infty}(\frac{1}{2}-\frac{1}{n+1})^n$$ which is less than your $\frac 1 2$ so it does go to $0$

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Another approach:

$$\left(\frac{n-1}{2n+2}\right)^n=\frac1{2^n}\left(1-\frac2{n+1}\right)^n=$$

$$=\frac1{2^n}\left(1-\frac2{n+1}\right)^{n+1}\left(1-\frac2{n+1}\right)^{-1}\xrightarrow[n\to\infty]{}0$$

since the second factor's limit above is $\;e^{-2}\;$ , and the third's is $\;1\;$

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You can calculate this limit easily by using the root test: First, we have to make sure that the sequence is bigger or equal to 0 for all n, i think it is quite easy to see it and then: $$a_n = \left(\frac{n-1}{2n+2}\right)^n$$ $$(a^n)^{1/n}=\frac{n-1}{2n+2} \rightarrow \frac{1}{2}$$ Therefore, by the root test: $$a_n \rightarrow 0$$ You can find about the root test here: http://en.wikipedia.org/wiki/Root_test, though in this link, it looks more complicated.

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