Limits of Sequences

A sequence is an infinite list of numbers $a_1,\ a_2,\ a_3,\ \ldots$ The term $a_n$ has the number (index) $n.$ The arithmetic and geometric progressions are examples of sequences.

Take the sequence ${a_n=\frac1n}$:

$1,\ \frac12,\ \frac13,\ \frac14,\ \ldots,$ $\frac{1}{100},\ \ldots,\ \frac{1}{1000},\ \ldots$

Its terms are always positive and never reach 0, but as $n$ grows they come as close to 0 as we like: $a_{100}$ differs from 0 by $0.01,$ $a_{1000}$ by $0.001,$ and so on. We say that the sequence tends to 0, or that its limit is 0, and write

This is read "the limit of $\frac1n$ as $n$ tends to infinity is 0". A shorter way to write it is ${\frac1n\to 0}.$

On this page: Definition · Infinite limits · Important limits · Limit rules · The squeeze theorem · How to find limits · Solved problems · Common mistakes · Practice problems

Definition of the limit

The inequality ${|a_n-a|\lt\varepsilon}$ means ${a-\varepsilon\lt a_n\lt a+\varepsilon}$: the term $a_n$ lies in the strip of width $2\varepsilon$ around $a.$ However narrow the strip is, from some index on all terms lie inside it. Only finitely many terms, the ones up to $n_\varepsilon,$ may stay outside.

${a_n=1+\frac{(-1)^{n+1}}{n}}$ and ${\varepsilon=0.25}$: every term after the fourth lies between ${1-\varepsilon}$ and ${1+\varepsilon},$ so here ${n_\varepsilon=4}.$

A sequence that has a limit (a number) is called convergent; otherwise it is divergent. A convergent sequence has only one limit, and it is bounded: all its terms lie between two fixed numbers.

Example. Prove that ${\lim_{n\to\infty}\frac{n+1}{n}=1}.$

Let ${a_n=\frac{n+1}{n}}.$ Then

${|a_n-1|}={\left|\frac{n+1}{n}-1\right|}={\frac1n}$

and ${\frac1n\lt\varepsilon}$ for every ${n\gt\frac1\varepsilon},$ so we can take ${n_\varepsilon=\frac1\varepsilon}.$ For example, for ${\varepsilon=0.01}$ all terms with ${n\gt 100}$ differ from 1 by less than 0.01.

The definition also shows that ${a_n\to a}$ means the same as ${a_n-a\to 0},$ and the same as ${|a_n-a|\to 0}.$

Infinite limits

The terms of ${a_n=n^2}$ grow without bound: $1,\ 4,\ 9,\ 16,\ \ldots$ For every number $M,$ however large, all terms from some index on are greater than $M.$ We then write ${\lim_{n\to\infty}n^2=+\infty}$ and say that the sequence tends to plus infinity. In the same way ${\lim_{n\to\infty}(-n)=-\infty}.$

A sequence with the limit $+\infty$ or $-\infty$ is still divergent, because $+\infty$ and $-\infty$ are not numbers. Some sequences have no limit at all, neither a number nor $\pm\infty.$ For example, ${b_n=(-1)^n}$ takes the values $-1,\ 1,\ -1,\ 1,\ \ldots$ and does not approach any one value.

Important limits

The terms of ${\left(1+\frac1n\right)^n}$ increase: ${2,\ 2.25,\ 2.370\ldots,\ 2.441\ldots}$ They never reach $e,$ and for every $n$

${\left(1+\frac1n\right)^n}\lt{e}\lt{\left(1+\frac1n\right)^{n+1}}$

so also ${\lim_{n\to\infty}\left(1+\frac1n\right)^{n+1}=e}.$ More generally, ${\lim_{n\to\infty}\left(1+\frac kn\right)^n=e^k}$ for every number $k.$ In particular, ${\lim_{n\to\infty}\left(1-\frac1n\right)^n=e^{-1}=\frac1e}.$

Limit rules

Let ${\lim_{n\to\infty}a_n=a}$ and ${\lim_{n\to\infty}b_n=b},$ where $a$ and $b$ are numbers. Then

In words: the limit of a sum, difference, product or quotient is the sum, difference, product or quotient of the limits, as long as both limits are numbers and we do not divide by 0.

Example.

${\lim_{n\to\infty}\left(3-\frac2n+\frac{5}{n^2}\right)}={3-2\cdot 0+5\cdot 0}={3}$

Rules for infinite limits

If ${a_n\to+\infty}$ and ${b_n\to+\infty},$ then

${a_n+b_n\to+\infty},$   ${a_n\cdot b_n\to+\infty},$   ${-a_n\to-\infty},$   ${\frac{1}{a_n}\to 0}$

${a_n^k\to+\infty}$ for ${k\gt 0},$   ${a_n^k\to 0}$ for ${k\lt 0}$

Also, if ${a_n\to+\infty}$ and $b_n$ has a limit that is a number, then ${a_n+b_n\to+\infty}.$ If ${a_n\to 0}$ and ${a_n\gt 0},$ then ${\frac{1}{a_n}\to+\infty}.$

The expressions ${\infty-\infty},$ ${\frac{\infty}{\infty}},$ ${0\cdot\infty},$ ${\frac00}$ and $1^\infty$ are indeterminate forms. The rules say nothing about them: such a limit can be any number, $\pm\infty,$ or may not exist. The sequence must first be transformed, as shown below.

Comparison and the squeeze theorem

Let ${\lim_{n\to\infty}a_n=a}$ and ${\lim_{n\to\infty}b_n=b}.$ If ${a_n\le b_n}$ for every $n,$ then ${a\le b}.$

Even if ${a_n\lt b_n}$ for every $n,$ we can only conclude ${a\le b},$ not ${a\lt b}.$ For example, ${0\lt\frac1n}$ for every $n,$ but both sequences have the limit 0.

Example. Since ${-1\le\sin n\le 1},$ we have ${-\frac1n\le\frac{\sin n}{n}\le\frac1n}.$ Both outer sequences tend to 0, so ${\lim_{n\to\infty}\frac{\sin n}{n}=0}.$

How to find limits

A quotient of polynomials in $n.$ Divide the numerator and the denominator by the highest power of $n$ in the denominator, then use ${\frac{1}{n^k}\to 0}$:

${\lim_{n\to\infty}\frac{2n^2-n+3}{5n^2+4}}={\lim_{n\to\infty}\frac{2-\frac1n+\frac{3}{n^2}}{5+\frac{4}{n^2}}}={\frac{2-0+0}{5+0}}={\frac25}$

In general, for polynomials $P(n)$ and $Q(n)$: if the degree of $P$ is less than the degree of $Q,$ then ${\frac{P(n)}{Q(n)}\to 0};$ if the degrees are equal, the limit is the quotient of the leading coefficients; if the degree of $P$ is greater, the limit is $+\infty$ or $-\infty.$

A difference of square roots (${\infty-\infty}$). Multiply and divide by the sum of the roots, so that ${(\sqrt A-\sqrt B)(\sqrt A+\sqrt B)}={{A-B}.}$ See problems 7 and 8.

Powers such as $2^n$ and $3^n.$ Divide by the power with the largest base and use ${q^n\to 0}$ for ${-1\lt q\lt 1}.$ See problem 10.

The form $1^\infty.$ Bring the sequence to the form ${\left(1+\frac1n\right)^n}$ or ${\left(1+\frac kn\right)^n}.$ See problem 9.

Solved problems

Problem 1. Find ${\lim_{n\to\infty}5\cdot\left(\frac14\right)^n}.$

Solution: Since ${-1\lt\frac14\lt 1},$ we have ${\left(\frac14\right)^n\to 0},$ and

${\lim_{n\to\infty}5\cdot\left(\frac14\right)^n}={5\cdot 0}={0}$

Compare: for ${5\cdot 4^n}$ the base is greater than 1, so ${5\cdot 4^n\to+\infty}.$

Problem 2. Find ${\lim_{n\to\infty}\frac{3n^2+1}{2n-n^2}}.$

Solution: We divide the numerator and the denominator by $n^2$:

${\lim_{n\to\infty}\frac{3n^2+1}{2n-n^2}}={\lim_{n\to\infty}\frac{3+\frac{1}{n^2}}{\frac2n-1}}={\frac{3+0}{0-1}}={-3}$

Problem 3. The sequence $(a_n)$ tends to 1, and ${a_n\ne 1}$ for every $n.$ Find ${\lim_{n\to\infty}\frac{2a_n^2-a_n-1}{a_n-1}}.$

Solution: The numerator and the denominator both tend to 0, so the rule for a quotient cannot be used directly. We factor the numerator: ${2a_n^2-a_n-1}={{(a_n-1)(2a_n+1)}.}$ Since ${a_n\ne 1},$ we can cancel ${a_n-1}$:

${\lim_{n\to\infty}\frac{2a_n^2-a_n-1}{a_n-1}}={\lim_{n\to\infty}\frac{(a_n-1)(2a_n+1)}{a_n-1}}={\lim_{n\to\infty}(2a_n+1)}={2\cdot 1+1}={3}$

Problem 4. Prove with the definition that ${\lim_{n\to\infty}\frac{2n+1}{n+3}=2}.$

Solution: Let ${a_n=\frac{2n+1}{n+3}}.$ Then

${|a_n-2|}={\left|\frac{2n+1}{n+3}-2\right|}={\left|\frac{2n+1-2n-6}{n+3}\right|}={\frac{5}{n+3}}$

and ${\frac{5}{n+3}\lt\varepsilon}$ for every ${n\gt\frac5\varepsilon-3}.$ So we can take ${n_\varepsilon=\frac5\varepsilon-3}.$ For example, for ${\varepsilon=0.1}$ we get ${n_\varepsilon=47}$: from the 48th term on, all terms differ from 2 by less than 0.1.

Problem 5. Find ${\lim_{n\to\infty}\frac{n^3-2n}{4n^3+n^2+5}}.$

Solution: We divide by $n^3$:

${\lim_{n\to\infty}\frac{n^3-2n}{4n^3+n^2+5}}={\lim_{n\to\infty}\frac{1-\frac{2}{n^2}}{4+\frac1n+\frac{5}{n^3}}}={\frac{1-0}{4+0+0}}={\frac14}$

Problem 6. Find a) ${\lim_{n\to\infty}\frac{2n^2+3}{n^3+1}};$   b) ${\lim_{n\to\infty}\frac{n^3+1}{2n^2+3}}.$

Solution: a) We divide by $n^3$:

${\lim_{n\to\infty}\frac{2n^2+3}{n^3+1}}={\lim_{n\to\infty}\frac{\frac2n+\frac{3}{n^3}}{1+\frac{1}{n^3}}}={\frac{0+0}{1+0}}={0}$

b) We divide by $n^2.$ The numerator tends to $+\infty$ and the denominator tends to 2, so

${\lim_{n\to\infty}\frac{n^3+1}{2n^2+3}}={\lim_{n\to\infty}\frac{n+\frac{1}{n^2}}{2+\frac{3}{n^2}}}={+\infty}$

Problem 7. Find ${\lim_{n\to\infty}(\sqrt{n+1}-\sqrt n)}.$

Solution: This is the form ${\infty-\infty}.$ We multiply and divide by ${\sqrt{n+1}+\sqrt n}$:

${\lim_{n\to\infty}(\sqrt{n+1}-\sqrt n)}={\lim_{n\to\infty}\frac{(n+1)-n}{\sqrt{n+1}+\sqrt n}}={\lim_{n\to\infty}\frac{1}{\sqrt{n+1}+\sqrt n}}={0}$

because the denominator tends to $+\infty.$

Problem 8. Find ${\lim_{n\to\infty}(\sqrt{n^2+n}-n)}.$

Solution: Again ${\infty-\infty}.$ We multiply and divide by ${\sqrt{n^2+n}+n},$ then divide by $n$:

${\lim_{n\to\infty}(\sqrt{n^2+n}-n)}={\lim_{n\to\infty}\frac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}}={\lim_{n\to\infty}\frac{n}{\sqrt{n^2+n}+n}}={\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac1n}+1}}={\frac{1}{1+1}}={\frac12}$

Problem 9. Find ${\lim_{n\to\infty}\left(1+\frac1n\right)^{3n}}.$

Solution: This is the form $1^\infty.$ We write the power as a cube:

${\lim_{n\to\infty}\left(1+\frac1n\right)^{3n}}={\lim_{n\to\infty}\left(\left(1+\frac1n\right)^n\right)^3}={e^3}$

Problem 10. Find ${\lim_{n\to\infty}\frac{2^n+3^n}{3^{n+1}-2^n}}.$

Solution: We divide the numerator and the denominator by $3^n$ and use ${\left(\frac23\right)^n\to 0}$:

${\lim_{n\to\infty}\frac{2^n+3^n}{3^{n+1}-2^n}}={\lim_{n\to\infty}\frac{\left(\frac23\right)^n+1}{3-\left(\frac23\right)^n}}={\frac{0+1}{3-0}}={\frac13}$

Common mistakes

Practice problems

Find the limits. Try first, then open the answer.

See also: Derivative formulas (the derivative is defined as a limit) and the function graph plotter.