×

# Limits of Sequences

## Definition

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:

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

## Technique

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

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

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

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

Here are some rules for the limits of sequences:

\begin{align} \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{align}

Now, look at these examples:

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

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

$\frac{12n^2-3n}{n^2+5n+9} = \frac{12-\frac{3}{n}}{1+\frac{5}{n}+\frac{9}{n^2}}$

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

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

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

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

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

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

Note by Arron Kau
3 years ago

Sort by:

can't we simply put y=1/x and take limit as y approaches zero · 2 years, 9 months ago