Limits of Sequences

10 min read

Textbook Reference

Listing the terms of a sequence and reading the notation $\{a_n\}$ come first; a limit asks where those terms head.

Key idea

You already know how to find $\displaystyle\lim_{x\to\infty} f(x)$ for an ordinary function. The limit of a sequence is the same computation with the inputs restricted to whole numbers.

Picture the graph of a sequence: a row of separated dots, one above each whole number $n$. As you read the dots left to right, ask one question. Do the heights settle toward a single level line, or not? If the dots crowd toward a height $L$ and stay near it, the sequence converges to $L$. If they wander, oscillate, or run off to infinity, the sequence diverges.

Here is the bridge that does almost all of the work. A sequence is a function whose inputs happen to be integers. So if the matching continuous function $f(x)$ has a limit $L$ as $x \to \infty$, then the sequence $a_n = f(n)$ has the same limit $L$. That single fact lets you reuse every limit technique you already know, including dividing by the highest power and l’Hopital’s rule, just by switching the variable from $n$ to a continuous $x$ when you need to.

Prerequisite Check

• List the terms of a sequence from a formula (the previous lesson)
• Find $\displaystyle\lim_{x\to\infty} f(x)$ for a rational function by dividing by the highest power of $x$
• Apply l’Hopital’s rule to an $\infty/\infty$ or $0/0$ limit of a function
• Know that $\displaystyle\lim_{x\to\infty} \frac{1}{x} = 0$
• Recognize when a value “does not exist” because terms oscillate

If these are solid, you are ready.

Quick Reference

Intuitive definition. $\displaystyle\lim_{n\to\infty} a_n = L$ means the terms $a_n$ can be made as close to $L$ as you like by taking $n$ large enough. If this limit exists, the sequence converges; otherwise it diverges.

Precise definition (formal gloss). $\displaystyle\lim_{n\to\infty} a_n = L$ means: for every $\varepsilon > 0$ there is an integer $N$ such that if $n > N$ then $|a_n - L| < \varepsilon$. (In plain words: name any error tolerance $\varepsilon$, and from some point $N$ onward every term lands within that tolerance of $L$.)

The function-limit bridge (Theorem 4). If $\displaystyle\lim_{x\to\infty} f(x) = L$ and $f(n) = a_n$ for integer $n$, then $\displaystyle\lim_{n\to\infty} a_n = L$. This is what lets you use l’Hopital’s rule on a related continuous function.

Two facts worth memorizing. \[ \lim_{n\to\infty} \frac{1}{n^r} = 0 \quad (r > 0), \qquad \lim_{n\to\infty} r^n = \begin{cases} 0 & \text{if } -1 < r < 1 \\ 1 & \text{if } r = 1 \end{cases} \] and $\{r^n\}$ diverges for all other $r$.

Limit Laws for sequences. If $\{a_n\}$ and $\{b_n\}$ converge, then sums, differences, constant multiples, products, and quotients of limits behave exactly as for functions (the quotient law needs the bottom limit nonzero).

Squeeze Theorem. If $a_n \le b_n \le c_n$ for large $n$ and $\lim a_n = \lim c_n = L$, then $\lim b_n = L$.

Absolute-value test (Theorem 6). If $\displaystyle\lim_{n\to\infty} |a_n| = 0$, then $\displaystyle\lim_{n\to\infty} a_n = 0$.

Key Concepts

1. What the Limit Means

Look again at the sequence $a_n = \dfrac{n}{n+1}$, whose first terms are $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \tfrac{4}{5}, \dots$. The terms climb toward $1$ but never reach it. The gap between the term and $1$ is \[ 1 - \frac{n}{n+1} = \frac{1}{n+1}, \] and that gap shrinks toward $0$ as $n$ grows. So the terms can be made as close to $1$ as you please. We write $\displaystyle\lim_{n\to\infty} \frac{n}{n+1} = 1$.

The precise definition turns “as close as you please” into a challenge-and-response. Someone names a tolerance $\varepsilon$ (think of a thin horizontal band of half-width $\varepsilon$ around the line $y = L$). You must produce a cutoff $N$ so that every dot past position $N$ lands inside the band. If you can always answer, no matter how thin the band, the limit is $L$. A smaller band usually forces a larger $N$, but as long as some $N$ always works, the sequence converges.

A sequence diverges when no single $L$ works. There are two flavors. The terms might run off to infinity (write $\lim a_n = \infty$ to record that), or they might oscillate forever between values without settling, as $\{(-1)^n\} = -1, 1, -1, 1, \dots$ does.

2. The Function-Limit Bridge

The reason sequence limits are not a whole new subject is Theorem 4. Compare the precise definition for a sequence with the one for $\lim_{x\to\infty} f(x)$; the only difference is that $n$ must be an integer. So if the continuous function agrees with the sequence at the integers and the function has a limit, the sequence inherits it.

Example 1 (divide by the highest power). Find $\displaystyle\lim_{n\to\infty} \frac{n}{n+1}$.

Goal. Use the same move as for a rational-function limit at infinity: divide top and bottom by the highest power of $n$ in the denominator.

Work. \[ \lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1 + \frac{1}{n}} = \frac{1}{1 + 0} = 1 \] using $\lim_{n\to\infty} \frac{1}{n} = 0$.

Answer: $1$

Recap. The technique is identical to the function case. The terms confirm it: $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \dots$ head to $1$.

Example 2 (when the bridge requires l’Hopital). Find $\displaystyle\lim_{n\to\infty} \frac{\ln n}{n}$.

Goal. Both numerator and denominator grow without bound, an $\infty/\infty$ form. l’Hopital’s rule is the tool, but it applies to functions of a real variable, not to sequences directly. So apply it to the related function and then use Theorem 4.

Work. Consider $f(x) = \dfrac{\ln x}{x}$. Then \[ \lim_{x\to\infty} \frac{\ln x}{x} = \lim_{x\to\infty} \frac{1/x}{1} = 0. \] Because $f(n) = \dfrac{\ln n}{n}$, Theorem 4 gives \[ \lim_{n\to\infty} \frac{\ln n}{n} = 0. \]

Answer: $0$

Recap. You cannot differentiate a sequence, because there is nothing between the integer inputs to take a derivative over. The fix is to pass to the continuous function, do l’Hopital there, then carry the answer back by Theorem 4.

(Common error: writing “$\frac{d}{dn}$” and applying l’Hopital to the sequence itself. l’Hopital’s rule is a statement about differentiable functions of a real variable. Always switch to the related $f(x)$ first, as Stewart does in this example.)

3. Limit Laws and the Squeeze Theorem

Because sequence limits behave like function limits, the Limit Laws carry over. Sums of limits, products of limits, and quotients of limits all work as expected, provided the pieces converge (and the denominator limit is nonzero for a quotient).

When a sequence is too tangled for the Limit Laws, the Squeeze Theorem often rescues it: trap the sequence between two simpler sequences that share a limit.

Example 3 (Squeeze Theorem). Discuss the convergence of $a_n = \dfrac{n!}{n^n}$, where $n! = 1 \cdot 2 \cdot 3 \cdots n$.

Goal. There is no continuous function for $x!$, so l’Hopital is out. Bound the sequence instead.

Work. Write the ratio as \[ a_n = \frac{1 \cdot 2 \cdot 3 \cdots n}{n \cdot n \cdot n \cdots n} = \frac{1}{n}\left( \frac{2 \cdot 3 \cdots n}{n \cdot n \cdots n} \right). \] The expression in parentheses is at most $1$, because each factor in its numerator is no larger than $n$. Therefore \[ 0 < a_n \le \frac{1}{n}. \] Since the lower bound $0$ and the upper bound $\dfrac{1}{n}$ both go to $0$, the Squeeze Theorem forces \[ \lim_{n\to\infty} \frac{n!}{n^n} = 0. \]

Answer: $0$ (the sequence converges)

Recap. When a sequence has no matching continuous function, look for a clean upper bound and a clean lower bound with the same limit. Here both bounds pinch to $0$.

4. Oscillation and the Absolute-Value Test

Example 4 (a divergent oscillation). Determine whether $a_n = (-1)^n$ converges.

Goal. List the terms and watch their behavior.

Work. The terms are $-1, 1, -1, 1, -1, \dots$. They jump between $-1$ and $1$ forever and never approach a single number.

Answer: the sequence diverges (the limit does not exist)

Recap. Bouncing between two values is not converging. There is no $L$ within $\varepsilon = \tfrac{1}{2}$ of every later term.

Example 5 (oscillation that does converge). Evaluate $\displaystyle\lim_{n\to\infty} \frac{(-1)^n}{n}$.

Goal. The sign flips, but the size shrinks. Use the absolute-value test (Theorem 6).

Work. First take the limit of the absolute value: \[ \lim_{n\to\infty} \left| \frac{(-1)^n}{n} \right| = \lim_{n\to\infty} \frac{1}{n} = 0. \] Because the absolute values go to $0$, Theorem 6 gives \[ \lim_{n\to\infty} \frac{(-1)^n}{n} = 0. \]

Answer: $0$

Recap. Alternating signs do not block convergence on their own. What matters is whether the terms shrink toward a single value. When the absolute values go to $0$, the original sequence goes to $0$ as well.

(Common error: declaring any alternating sequence divergent. The sign is only half the story. $\{(-1)^n\}$ diverges because its size stays at $1$; $\{(-1)^n/n\}$ converges because its size goes to $0$.)

Inline Self-Check

Question. Without computing a derivative, find $\displaystyle\lim_{n\to\infty} \frac{4n^2 - 3n}{2n^2 + 1}$.

Common Errors Summary

Common Misconceptions

Leveled Practice

Attempt each problem before opening the answer.

Level 1: Direct Application

Problem 1. Find $\displaystyle\lim_{n\to\infty} \frac{5}{n+2}$.

Problem 2. Find $\displaystyle\lim_{n\to\infty} \frac{3\sqrt{n}}{\sqrt{n+2}}$.

Problem 3. Determine whether $a_n = 2 + (0.86)^n$ converges, and if so, to what.

Level 2: Bridge and Squeeze

Problem 4. Find $\displaystyle\lim_{n\to\infty} \sin\frac{\pi}{n}$.

Problem 5. Evaluate $\displaystyle\lim_{n\to\infty} \frac{(\ln n)^2}{n}$.

Problem 6. Use the Squeeze Theorem to find $\displaystyle\lim_{n\to\infty} \frac{\cos^2 n}{2^n}$.

Level 3: Deeper Problems

Problem 7. For which values of $r$ does $\{r^n\}$ converge, and what is the limit in each case?

Problem 8. Explain, using the precise definition, what cutoff $N$ guarantees that every term of $a_n = \dfrac{1}{n}$ past position $N$ is within $\varepsilon = 0.01$ of its limit $0$.

Mastery Checklist

You have mastered this skill when you can do all of the following without notes:

• State the intuitive meaning of $\lim_{n\to\infty} a_n = L$
• State the precise definition in challenge-and-response form
• Decide whether a sequence converges or diverges from its terms
• Use the function-limit bridge (Theorem 4) to apply l’Hopital correctly
• Apply the Limit Laws to sums, products, and quotients of convergent sequences
• Use the Squeeze Theorem when there is no matching continuous function
• Use the absolute-value test on a shrinking alternating sequence
• Recall the limits of $1/n^r$ and $r^n$

Mental Model

Keep the row-of-dots picture from the previous lesson, then add a thin horizontal band around a candidate level line $y = L$. Convergence is a game. Your opponent makes the band thinner and thinner. You win if, for every band, you can name a position $N$ past which every dot lands inside the band. Win every round and the sequence converges to $L$. If your opponent can find a band you cannot satisfy, no matter how far out you push $N$, then $L$ is not the limit (and if no $L$ works, the sequence diverges).

This picture also explains the bridge to functions. If you draw the smooth curve $y = f(x)$ through the dots and the curve flattens toward $L$, the dots sitting on the curve flatten toward $L$ too. That is Theorem 4 in one sentence, and it is why every function-limit trick is available to you.

Connections

Built From

• Sequences and notation: you must be able to read $\{a_n\}$ and list terms before you can ask where they head.
• Limits of functions at infinity and l’Hopital’s rule: the entire toolkit transfers through Theorem 4.

Leads To

• Properties of sequences: monotonic and bounded sequences, and the theorem that a bounded monotonic sequence must converge even when you cannot compute the limit directly.
• Infinite series and the divergence test: a series can converge only if its terms shrink to $0$, which is a limit-of-a-sequence question. The divergence test is the immediate payoff of this lesson.

Why It Matters

A limit of a sequence is the precise meaning of “this process settles down.” Newton’s method converges when the sequence of approximations has a limit equal to the root. A numerical algorithm is trustworthy exactly when the sequence of its outputs converges, and the cutoff $N$ from the precise definition is the iteration count that achieves a target accuracy $\varepsilon$. The challenge-and-response definition is not abstract decoration; it is how error tolerances are specified in scientific computing.

Audience Notes

For students who find math intimidating: you already did this for functions. The only new instruction is to restrict the inputs to whole numbers, and then most problems are solved by techniques you have used before.

For students who want depth: the precise $\varepsilon$-$N$ definition is the first rigorous limit definition many students meet on its own terms. Sitting with the challenge-and-response game here pays off later, because the definition of a limit of a function and the convergence of a series both run on the same logic.

For students aimed at a career in computing or engineering: convergence with a known $N$ for a given $\varepsilon$ is a stopping criterion. Iterative solvers, from root-finders to machine-learning optimizers, run until the sequence of estimates is within tolerance, which is this definition put to work.

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