# Does $\lim \limits_ {n\to \infty}\ \frac{n}{n+1}$ equal $1$?

$$\lim_ {n\to \infty}\ \frac{n}{n+1}=1$$

I'd write it like this:

$$\lim_ {n\to \infty}\ \frac{n}{n}+\frac{n}{1}=\lim_ {n\to \infty}\ 1+n=\infty$$

What am I missing?

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Your algebra is wrong! – The Chaz 2.0 Nov 18 '11 at 19:47
1) $x$ and $n$ are different variables. 2) Division doesn't work like that. – Qiaochu Yuan Nov 18 '11 at 19:47
divide both numerator and denominator by $n$ – pedja Nov 18 '11 at 19:48
Note that $\frac{n}{n+1}\neq\frac{n}{n}+\frac{n}{1}$. But it worth use equality $\frac{n}{n+1}=\frac{1}{1+\frac{1}{n}}$ – Norbert Nov 18 '11 at 19:49
You should realize that your algebra is wrong: $\frac{n}{n+1} <1$ while $\frac{n}{n}+\frac{n}{1} >1$... How can those be equal? – N. S. Nov 18 '11 at 20:11

Since no one has said so explicitly, to handle a limit of this form (taking a limit at infinity of a rational expression), start of by dividing every additive term by the highest power of $n$ that you see. More precisely, multiply both top and bottom by the reciprocal of the highest power of $n$ that you see; which, in this case is just $n$:

$$\lim_{n\rightarrow\infty}{n\over n+1}= \lim_{n\rightarrow\infty}\Bigl[\underbrace{{1/ n}\over{1/ n}}_{=1}\cdot{n\over n+1} \Bigr] = \lim_{n\rightarrow\infty}{ {1\over n} \cdot n \over {{1\over n} (n+1)}} = \lim_{n\rightarrow\infty}{ 1 \over {1+{1\over n} }} ={1\over 1+0}=1.$$

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What you are missing is that, in general, $$\frac{a}{b+c}\neq\frac{a}{b}+\frac{a}{c}.$$ That is, your two functions are not equal.

Note that $\frac{n}{n}+\frac{n}{1} = 1+n$, whereas $\frac{n}{n+1}\lt 1$ for all $n\gt 0$.

What you can do is to instead consider the reciprocal: $$\lim_{n\to\infty}\frac{n}{n+1} = \lim_{n\to\infty}\frac{1}{\quad\frac{n+1}{n}\quad} = \lim_{n\to\infty}\frac{1}{\frac{n}{n}+\frac{1}{n}} = \lim_{n\to\infty}\frac{1}{1+\frac{1}{n}}.$$ Now notice that $\frac{1}{n}\to 0$ as $n\to\infty$.

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You can better transform your limit, doing:

n+1 = x, so x-1=n, then your new limit pass from:

$$\lim_ {n\to \infty}\ \frac{n}{n+1}=1$$

to:

$$\lim_ {x\to \infty}\ \frac{x-1}{x}=\lim_ {x\to \infty}\ \frac{x}{x}-\frac{1}{x}=1+0=1$$

Yout problem is in the way you separate the terms.

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Yet another way...

$\frac{n}{n+1}$

can be written as:

$\frac{n+(1-1)}{n+1}$

which is:

$\frac{n+1}{n+1} -\frac{1}{n+1} = 1 - \frac{1}{n+1}$

so

$\lim_ {n\to \infty}\ \frac{n}{n+1} =\lim_ {n\to \infty}\ (1 - \frac{1}{n+1})$

since limit of 1/(n+1) as n approaches infinity is 0, and since the limit of 1 as n approaches infinity is 1, then the desired limit is = $1 - 0 =1$

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