We call $l$ the limit of a sequence $(x_n)$ if the following condition holds:
For any real number $\epsilon \gt 0$ there exists a natural number $N$ such that, for every $n\gt N$, we have $\lvert x_n - l \rvert \lt \epsilon$.
In other words, for every measure of closeness $\epsilon$, the sequence's terms are eventually closer than $\epsilon$ to the limit. The sequence $(x_n)$ is said to converge, or “tend to the limit $l$”. This relationship is expressed in notation as \[x_n \to l \; \text{ as } \; n \to \infty\] or \[\lim_{n \to \infty}x_n = l\]
It is often important to examine what happens to a sequence as $n$ gets very large. There are two types of behaviour:
A sequence $(x_n)$ tends to (positive) infinity, if, however large a number we choose, the sequence will eventually become greater than that number and stay greater. This can be denoted by \[x_n \to \infty \; \text{ as } \; n \to \infty\] or \[\lim_{n \to \infty}(x_n) = \infty\]
Example: the sequence $\bigl(n^2\bigr)^{\infty}_{n=1} \longrightarrow \infty$ as $n \to \infty$.
A sequence $(x_n)$ tends to negative infinity, if, however large a negative number we choose, the terms of the sequence will eventually become less than that number and stay less. This can be denoted by \[x_n \to -\infty \; \text{ as } \; n \to \infty\] or \[\lim_{n \to \infty}(x_n) = -\infty.\]
Example: the sequence $\bigl(-n^3\bigr)^{\infty}_{n=1} \longrightarrow -\infty$ as $n \to \infty$.
A sequence $(x_n)$ tends to a real limit $l$ if, however small an interval around $l$ we choose, the sequence will eventually take values within that interval, and stay there. This can be denoted by \[x_n \to l \; \text{ as } \; n \to \infty\] or \[\lim_{n\rightarrow\infty}(x_n) = l\]
Example: $\displaystyle{ \lim_{n \to \infty} \left(\frac{1}{n}\right) = 0 }$.
Does the following sequence tend to infinity, minus infinity or a real limit: $\bigl(2^n\bigr)^{\infty}_{n=1}$?
Write out the first few terms to get an idea of the behaviour.
\[2, 4, 8, 16, 32, 64, \dotso\]
These are increasing and appear to be tending towards infinity.
Check by picking a large number, say $1000$, and seeing if there are values of $n$ which give greater numbers than this in the sequence.
When $n=10$, $u_{10} = 1024$, and the next few terms are $2048, 4096, 8192, \dotso$ which stay greater than our chosen number, so
\[(2^n) \rightarrow \infty \text{ as } n\rightarrow\infty.\]
Does the following sequence tend to infinity, minus infinity or a real limit: $\bigl(1000-n\bigr)^{\infty}_{n=1}$?
Write out the first few terms to get an idea of the behaviour.
\[999, 998, 997, 996, 995, 994, \dotso\]
These are decreasing and appear to be tending towards minus infinity.
Check by picking a large negative number, say $-1000$, and seeing if there are values of $n$ which give smaller numbers than this in the sequence.
When $n=2001$, $u_{2001} = -1000$, and the next few terms are $-1001, -1002, -1003, \dotso$ which stay smaller than our chosen number, so
\[(1000-n) \rightarrow -\infty \text{ as } n\rightarrow\infty.\]
Does the following sequence tend to infinity, minus infinity or a real limit: $\left(5-\dfrac{1}{n}\right)^{\infty}_{n=1}$?
Write out the first few terms to get an idea of the behaviour.
\[4, 5-\frac{1}{2}, 5-\frac{1}{3}, 5-\frac{1}{4}, 5-\frac{1}{5}, \dotso\]
The fraction we are taking away each time is getting smaller and smaller.
We know that as $n$ gets larger, $\dfrac{1}{n}$ gets smaller and approaches zero, so as the terms go on it will almost be like $5-0$, therefore
\[\left(5-\dfrac{1}{n}\right) \rightarrow 5 \text{ as } n\rightarrow\infty.\]
Does the following sequence tend to infinity, minus infinity or a real limit: $\left(\sin\dfrac{\pi n}{4}\right)^{\infty}_{n=1}$?
Write out the first few terms to get an idea of the behaviour.
\[\frac{1}{ \sqrt{2} }, 1, \frac{1}{ \sqrt{2} }, 0, -\frac{1}{ \sqrt{2} }, -1, -\frac{1}{ \sqrt{2} }, 0, \frac{1}{ \sqrt{2} }, \dotso\]
This sequence of numbers increases then decreases and after eight terms starts to repeat.
Therefore this sequence has no limit.
Given the sequence $(x_n)_{n=1}^{\infty}$ where $x_n = \dfrac{8n-2}{13n+2}$. Find the least $N$ such that all terms from the $N$th are within $10^{-3}$ of the limit.
First find the limit of the sequence.
As $n\to \infty$, $8n$ and $13n$ will get so large that the constant term in both the numerator and denominator will have little effect. So we can think of this sequence as: \[x_n = \frac{8n}{13n}\] The $n$'s will always be the same no matter how large $n$ gets, so these will cancel with each other, therefore we have: \[\lim_{n\to\infty} \left(\frac{8n-2}{13n+2}\right) = \frac{8}{13}\]
Now, using the formula $\vert x_n - l \vert = \epsilon$, with $x_n = \dfrac{8n-2}{13n+2}, l=\dfrac{8}{13}$ and $\epsilon = 10^{-3}$. \begin{align} \left\vert \frac{8n-2}{13n+2} - \frac{8}{13} \right\vert &\leq 10^{-3}\\ \left\vert \frac{13(8n-2) -8(13n+2)}{13(13n+2)} \right\vert &\leq 10^{-3}\\ \frac{\vert 104n-26 -104n-16\vert}{169n+26} &\leq 10^{-3}\\ \frac{\vert -42 \vert}{169n+26}&\leq 10^{-3}\\ \end{align} We can take the absolute value signs off the denominator as it has to be positive so that the fraction is not undefined. Then rearrange and solve for $n$. \begin{align} 42 \times 10^3 &\leq 169n +26\\ 42000 &\leq 169n +26\\ 42000-26 &\leq 169n\\ n &\geq \frac{1}{169}(42000-26)\\ &\approx248.3669 \end{align} Hence the least integer value is given by rounding up: $\therefore N=249$.
Test yourself: Numbas test on limits of sequences