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Given the problem: Determine the limits of the sequnce $\{x_n\}^ \infty_{ n=1}$ $$x_n = \frac{n}{n+1}$$

The solution to this is:


$\lim\limits_{n \rightarrow \infty} x_n = \lim\limits_{n \rightarrow \infty} \frac{n}{n + 1}$


$=\lim\limits_{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}$


$=\frac{1}{1 + \lim\limits_{n \rightarrow \infty} \frac{1}{n}}$


$=\frac{1}{1 + 0}$



I get how you go from step 2 to 5 but I don't understand how you go from step 1 to 2.

Again, I'm stuck on the basic highschool math.

Please help

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multiply $n/(n+1)$ top and bottom by $1/n$, using the distributive law in the denominator. – Gerry Myerson Apr 19 '12 at 6:38
It is interesting to think what could the teacher have done for a student to understand limits but slip in the algebra. – Pedro Tamaroff Apr 21 '12 at 21:38
up vote 4 down vote accepted

Divide the numerator and denominator by $n$. Why is this legal, in other words, why does this leave your fraction unchanged?

Because $$\frac {\frac a n} {\frac b n}=\frac {a \cdot \frac 1 n} {b \cdot \frac 1 n}=\frac a b$$ where the last equality is because $\dfrac 1 n$'s get cancelled.

Further, remember the fact that:

$$\frac{a+b}{n}=\frac a n+\frac b n$$

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I like the simplicity of this approach. – Gineer Apr 20 '12 at 5:28
how do you decide that in this case you need to divide numerator and denominator by n what was the process behind this decision ? – user29654 Apr 22 '12 at 0:11

This is just algebraic manipulation from step 1 to step 2. Since $n \neq 0$, we can do the following. I will write it out in full detail so that you are clear on the steps involved.

$$\begin{eqnarray*} \frac{n}{n+1} &=& n \left(\frac{1}{n+1}\right)\\ &=& (n^{-1})^{-1} \left(\frac{1}{n+1}\right)\\ &=& \left(\frac{1}{n}\right)^{-1} \left(\frac{1}{n+1}\right) \\ &=&\frac{1}{\left(\frac{1}{n}\right)}\left(\frac{1}{n+1}\right)\\ &=& \frac{1}{ \left(\frac{1}{n}\right)(n+1)}\\ &=& \frac{1}{\left(\frac{n+1}{n}\right)} \\ &=& \frac{1}{\left( \frac{n}{n} + \frac{1}{n} \right)}\\ &=& \frac{1}{ \left( 1 + \frac{1}{n} \right)} \end{eqnarray*}$$

as desired.

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I like the way this approach works with the existing equation – Gineer Apr 20 '12 at 5:29

You can also do this way, $$\begin{eqnarray*} \lim_ {n\to \infty}x_n &=& \lim_{n\to \infty} \frac{n+1-1}{n+1}\\ &=& \lim_{n\to \infty}1-\frac{1}{n+1}\\ &=& 1-\lim_{n\to \infty}\frac{1}{n+1}\\ &=& 1 \end{eqnarray*}$$

because as $n \to \infty$, we have $\frac{1}{n+1} \to 0 .$

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