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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:

step1:

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

step2:

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

step3:

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

step4:

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

step5:

$=1$

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

  • 0
    It is interesting to think what could the teacher have done for a student to understand limits but slip in the algebra.2012-04-21

3 Answers 3

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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:

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

  • 0
    how do you decide that in this case you need to divide numerator and denominator by n what was the process behind this decision ?2012-04-22
4

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.

  • 0
    I like the way this approach works with the existing equation2012-04-20
1

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 $\frac{1}{n+1} \to 0 .$