Limits of Sequences


In a general sense, the limit of a sequence is a value that it approaches with arbitrary closeness. Stated precisely: \( \displaystyle \lim_{n \to \infty } \left\{ a_n \right\}=L \) means that for every \( \epsilon > 0 \) there exists a positive integer \( M \) so that:

If n>M, then anL<ϵ \text{If } n > M \text{, then } \left| a_n - L \right| < \epsilon


To take a simple example, let's find limn{1n}\displaystyle \lim_{n \to \infty }\left\{ \frac{1}{n} \right\} .

It's fairly obvious that as n n gets larger, the value of 1n \frac{1}{n} approaches 0, but how can we use the definition to help make that explicit?

Let M=1ϵ M = \lfloor \frac{1}{\epsilon} \rfloor . Then for any n>M n>M :

1n0<11/ϵ0ϵ. \left| \frac{1}{n} -0 \right| < \left| \frac{1}{\lfloor 1/\epsilon\rfloor} -0 \right| \leq \epsilon.

Here are some rules for the limits of sequences:

limn(an±bn)=limnan±limnbnlimn(can)=climnanlimn(anbn)=(limnan)(limnbn)limnanbn=limnanlimnbn, as long as limnbn0 \begin{aligned} \lim_{n \to \infty}\left( a_n \pm b_n \right) &= \lim_{n \to \infty} a_n \pm \lim_{n \to \infty} b_n \\ \lim_{n \to \infty} (c\cdot a_n) &= c \cdot \lim_{n \to \infty} a_n \\ \lim_{n \to \infty} \left( a_n b_n \right) &= \left( \lim_{n \to \infty} a_n \right) \left(\lim_{n \to \infty} b_n \right) \\ \lim_{n \to \infty} \frac{a_n}{b_n} &= \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} \text{, as long as } \lim_{n \to \infty} b_n \neq 0 \end{aligned}

Now, look at these examples:

What is limnan \displaystyle \lim_{n \to \infty} a_n if {an}=12n23nn2+5n+9\displaystyle \left\{ a_n\right\} = \frac{12n^2-3n}{n^2+5n+9} ?

First, we can re-express the limit by dividing both numerator and denominator by n2 n^2 :

12n23nn2+5n+9=123n1+5n+9n2 \frac{12n^2-3n}{n^2+5n+9} = \frac{12-\frac{3}{n}}{1+\frac{5}{n}+\frac{9}{n^2}}

Using the above rules, we can see that limnan=1201+0+0=12 \displaystyle \lim_{n \to \infty} a_n = \frac{12-0}{1+0+0} = 12 . _\square


Find limn31/n \displaystyle \lim_{n \to \infty} 3^{1/n} .

Since 1n+1<1n \frac{1}{n+1}< \frac{1}{n} , it is clear that 31/(n+1)<31/n 3^{1/(n+1)}<3^{1/n} , which means the sequence must be decreasing as n n increases. Further, it is apparent that 1n3 1^n \leq 3 for any value of n n , so it follows that 131/n 1 \leq 3^{1/n} . This gives us limn31/n=L \lim_{n \to \infty} 3^{1/n} = L such that:

1L31/n for any value of n>1. 1 \leq L \leq 3^{1/n} \text{ for any value of } n > 1.

The limit must then be 1 1 . To see why, assume that it is not. Then we have: 1nLn3 as n. 1^n \leq L^n \leq 3 \text{ as } n \to \infty.

Since this is clearly absurd for any L>1 L > 1 , we see that the limit must be 1 1 . _\square

Note by Arron Kau
7 years, 4 months ago

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can't we simply put y=1/x and take limit as y approaches zero

nishant singh - 7 years, 1 month ago

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Of course, but he's just showing a different way of looking/tackling it.

Milly Choochoo - 7 years, 1 month ago

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