Derivatives: rules and functions

Limits of sequences

The limit of a sequence is the value its general term approaches as nn grows without bound; for a fraction, divide the numerator and denominator by the highest power of nn in the denominator.

lim⁡n→∞1n=0,lim⁡n→∞rn=∞(r>1)\begin{gathered} \lim_{n\to\infty}\frac{1}{n}=0, \\[4pt] \lim_{n\to\infty}r^n=\infty\quad(r>1) \end{gathered}

As n→∞n\to\infty, 1n\frac{1}{n} and 1n2\frac{1}{n^2} approach 00. Dividing the numerator and denominator of a fractional sequence by the highest power in the denominator produces these terms, and the limit can then be read off.

lim⁡n→∞2n2+n3n2−3=lim⁡n→∞2+1n3−3n2=23\begin{gathered} \lim_{n\to\infty}\frac{2n^2+n}{3n^2-3} \\[4pt] =\lim_{n\to\infty}\frac{2+\frac{1}{n}}{3-\frac{3}{n^2}} \\[4pt] =\frac{2}{3} \end{gathered}

If the numerator and denominator have the same degree, the limit is the ratio of the leading coefficients. A sequence that grows without bound, such as rnr^n with r>1r>1, diverges to positive infinity.

Examples

  1. 01What is the value of this limit?lim⁡n→∞2n2+n3n2−3\displaystyle \lim_{n \to \infty} \frac{2n^{2}+n}{3n^{2}-3}AnswerHide
    Answer23\displaystyle \frac{2}{3}

    Steps

    1. 2n2+n3n2−3\displaystyle \frac{2n^{2}+n}{3n^{2}-3}=2+1n3−3n2\displaystyle = \frac{2+\frac{1}{n}}{3-\frac{3}{n^{2}}}
    2. As n grows the leftover terms vanish
    3. lim⁡n→∞2n2+n3n2−3\displaystyle \lim_{n \to \infty} \frac{2n^{2}+n}{3n^{2}-3}=23\displaystyle = \frac{2}{3}
  2. 02Dividing the numerator and denominator by n2n^2 gives this expression. What is its value as n→∞n\to\infty?4+3n2−4n2\displaystyle \frac{4+\frac{3}{n}}{2-\frac{4}{n^{2}}}AnswerHide
    Answer2\displaystyle 2

    Steps

    1. 3n\displaystyle \frac{3}{n}→0\displaystyle \to 04n2\displaystyle \frac{4}{n^{2}}→0\displaystyle \to 0
    2. Only the coefficients remain
    3. 4+02−0\displaystyle \frac{4+0}{2-0}=2\displaystyle = 2
  3. 03What is the value of this limit?lim⁡n→∞6n2+2n2n2−2\displaystyle \lim_{n \to \infty} \frac{6n^{2}+2n}{2n^{2}-2}AnswerHide
    Answer3\displaystyle 3

    Steps

    1. 6n2+2n2n2−2\displaystyle \frac{6n^{2}+2n}{2n^{2}-2}=6+2n2−2n2\displaystyle = \frac{6+\frac{2}{n}}{2-\frac{2}{n^{2}}}
    2. As n grows the leftover terms vanish
    3. lim⁡n→∞6n2+2n2n2−2\displaystyle \lim_{n \to \infty} \frac{6n^{2}+2n}{2n^{2}-2}=3\displaystyle = 3
  4. 04What is the value of this limit?lim⁡n→∞3n\displaystyle \lim_{n \to \infty} \frac{3}{n}AnswerHide
    Answer0\displaystyle 0

    Steps

    1. n=10\displaystyle n = 10⇒310\displaystyle \Rightarrow \frac{3}{10}n=100\displaystyle n = 100⇒3100\displaystyle \Rightarrow \frac{3}{100}
    2. The top stays 3 while the bottom grows
    3. lim⁡n→∞3n\displaystyle \lim_{n \to \infty} \frac{3}{n}=0\displaystyle = 0
  5. 05What is the value of this limit?lim⁡n→∞6n2+4n4n2−1\displaystyle \lim_{n \to \infty} \frac{6n^{2}+4n}{4n^{2}-1}AnswerHide
    Answer32\displaystyle \frac{3}{2}

    Steps

    1. 6n2+4n4n2−1\displaystyle \frac{6n^{2}+4n}{4n^{2}-1}=6+4n4−1n2\displaystyle = \frac{6+\frac{4}{n}}{4-\frac{1}{n^{2}}}
    2. As n grows the leftover terms vanish
    3. lim⁡n→∞6n2+4n4n2−1\displaystyle \lim_{n \to \infty} \frac{6n^{2}+4n}{4n^{2}-1}=32\displaystyle = \frac{3}{2}
  6. 06What is the value of this limit?lim⁡n→∞2n2+2n5n2−1\displaystyle \lim_{n \to \infty} \frac{2n^{2}+2n}{5n^{2}-1}AnswerHide
    Answer25\displaystyle \frac{2}{5}

    Steps

    1. 2n2+2n5n2−1\displaystyle \frac{2n^{2}+2n}{5n^{2}-1}=2+2n5−1n2\displaystyle = \frac{2+\frac{2}{n}}{5-\frac{1}{n^{2}}}
    2. As n grows the leftover terms vanish
    3. lim⁡n→∞2n2+2n5n2−1\displaystyle \lim_{n \to \infty} \frac{2n^{2}+2n}{5n^{2}-1}=25\displaystyle = \frac{2}{5}

FAQ

Q1Should I divide by nn or by n2n^2?AnswerHide
A
Divide by the highest power in the denominator. For 6n+14n−4\frac{6n+1}{4n-4}, divide by nn to get 6+1n4−4n\frac{6+\frac{1}{n}}{4-\frac{4}{n}}, which converges to 64=32\frac{6}{4}=\frac{3}{2}.
Q2If a sequence diverges to ∞\infty, is its limit ∞\infty?AnswerHide
A
∞\infty is not a number, so the sequence has no limit value. We write lim⁡n→∞an=∞\lim_{n\to\infty}a_n=\infty and say it diverges to positive infinity.

Published 2026-10-14 · Updated 2026-10-14