Limits of Sequences and Functions

Calculus begins with a single question dressed up in different costumes: what happens to a quantity as something else gets arbitrarily close to a particular value? Answering that question precisely — not just "it looks like it heads towards..." but with an argument that survives scrutiny — is the job of the limit. Every idea in the rest of this block, from the derivative to the definite integral, is built on the limit concept developed here, so it is worth being unhurried about it now.

To follow this chapter you should be comfortable with algebraic manipulation (factoring, rationalising surds), the graphs of polynomial and rational functions, and interval notation. Nothing here requires calculus itself — limits come before the derivative, not after.

1. The Limit of a Sequence

A sequence $(a_n)$ is a list of numbers indexed by the positive integers: $a_1, a_2, a_3, \ldots$. Consider $a_n = \dfrac{1}{n}$:

As $n$ grows without bound, $a_n$ gets closer and closer to $0$ — not reaching it for any finite $n$, but getting closer than any tolerance you name, provided $n$ is large enough. We write

Definition. A sequence $(a_n)$ has limit $L$, written $\displaystyle\lim_{n\to\infty} a_n = L$, if $a_n$ can be made arbitrarily close to $L$ by taking $n$ sufficiently large. If such an $L$ exists, the sequence is said to converge; otherwise it diverges.

The sequence $a_n = (-1)^n$ diverges: it never settles near one value, oscillating forever between $-1$ and $1$. The sequence $a_n = n$ also diverges, but in the specific way of growing without bound; we write $\displaystyle\lim_{n\to\infty} n = \infty$ while still calling it divergent, since $\infty$ is not a real number the sequence actually reaches.

2. The Limit of a Function at a Point

The same idea transfers to functions, but now the input variable $x$ can approach a real number $a$ rather than just increasing through the integers.

Definition. $\displaystyle\lim_{x\to a} f(x) = L$ means that $f(x)$ gets arbitrarily close to $L$ as $x$ gets arbitrarily close to $a$, from either side, without $x$ ever equalling $a$ itself.

That last clause matters enormously: the limit describes the behaviour approaching $a$, and is completely indifferent to whatever value $f(a)$ actually takes — or whether $f(a)$ is even defined. Consider

which is undefined at $x = 2$ (division by zero). Yet for every $x \neq 2$,

so as $x \to 2$, $f(x) \to 4$. We write $\displaystyle\lim_{x\to 2} f(x) = 4$ even though $f(2)$ does not exist. The graph of $f$ is the line $y = x+2$ with a single point removed at $(2,4)$ — a hole — and the limit correctly reports what the function is "trying to do" there.

3. One-Sided Limits and Existence of the Limit

Sometimes a function behaves differently depending on which side $x$ approaches from. The left-hand limit $\displaystyle\lim_{x\to a^-} f(x)$ considers only $x < a$ approaching $a$; the right-hand limit $\displaystyle\lim_{x\to a^+} f(x)$ considers only $x > a$.

Key distinction. The two-sided limit $\displaystyle\lim_{x\to a} f(x)$ exists if and only if both one-sided limits exist and are equal. If they disagree, the two-sided limit does not exist at all — it is not "one of the two values", it simply fails to exist.

Worked example. Let $f(x) = \dfrac{|x|}{x}$ for $x \neq 0$. For $x > 0$, $|x| = x$, so $f(x) = 1$; for $x < 0$, $|x| = -x$, so $f(x) = -1$. Hence

Since $-1 \neq 1$, $\displaystyle\lim_{x\to 0} f(x)$ does not exist, despite each one-sided limit existing perfectly well.

4. The Algebra of Limits

Provided the individual limits exist, limits interact with the arithmetic operations exactly as you would hope. If $\displaystyle\lim_{x\to a} f(x) = L$ and $\displaystyle\lim_{x\to a} g(x) = M$, then:

For any polynomial $p(x)$, this immediately gives $\displaystyle\lim_{x\to a} p(x) = p(a)$: just substitute. The same holds for a rational function $\dfrac{p(x)}{q(x)}$ at any $a$ with $q(a) \neq 0$. Limits only become interesting — and require technique — when direct substitution produces an indeterminate form such as $\dfrac{0}{0}$.

5. Resolving the $\frac{0}{0}$ Form

When substitution gives $\dfrac{0}{0}$, it means the expression has a common factor that cancels, exactly as in the worked example of Section 2. The method is always: factor, cancel, then substitute.

Worked example. Evaluate $\displaystyle\lim_{x\to 3} \frac{x^2 - 2x - 3}{x^2 - 9}$.

Direct substitution gives $\dfrac{9 - 6 - 3}{9-9} = \dfrac{0}{0}$, so both numerator and denominator share a factor of $(x-3)$:

Now substituting is valid:

When the indeterminate form involves a surd instead of a polynomial factor, rationalise the numerator or denominator by multiplying by the conjugate. For instance, $\displaystyle\lim_{x\to 0} \frac{\sqrt{x+4}-2}{x}$ is resolved by multiplying top and bottom by $\sqrt{x+4}+2$, turning the numerator into $(x+4)-4 = x$, which cancels the $x$ in the denominator and leaves $\dfrac{1}{\sqrt{x+4}+2} \to \dfrac{1}{4}$.

6. Limits at Infinity of Rational Functions

For $\displaystyle\lim_{x\to\infty} \frac{p(x)}{q(x)}$ with $p, q$ polynomials, divide every term by the highest power of $x$ appearing in the denominator; every term of the form $\dfrac{c}{x^k}$ then tends to $0$.

Worked example. Evaluate $\displaystyle\lim_{x\to\infty} \frac{3x^2 - 5x + 1}{2x^2 + 7}$.

The highest power in the denominator is $x^2$. Dividing every term, top and bottom, by $x^2$:

The general rule falls out of the same method: comparing the degrees of $p$ and $q$,

This result is the computational heart of the next chapter, where it identifies horizontal asymptotes.

7. The Squeeze (Sandwich) Theorem

Some limits resist both factoring and direct substitution, yet can be pinned down by trapping the function between two others whose limits agree.

Squeeze theorem. If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$ (possibly excluding $a$), and $\displaystyle\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L$, then $\displaystyle\lim_{x\to a} f(x) = L$ too — $f$ is squeezed into the same limit.

The standard application is $\displaystyle\lim_{x\to 0} \frac{\sin x}{x} = 1$ (with $x$ in radians). A geometric comparison of the areas of a small triangle, a circular sector and a larger triangle built on a unit circle shows that $\cos x \leq \dfrac{\sin x}{x} \leq 1$ for small $x \neq 0$. Since $\cos x \to 1$ and the constant function $1 \to 1$ as $x \to 0$, the sandwiched quantity $\dfrac{\sin x}{x}$ is forced to the same limit, $1$. This single result becomes indispensable once trigonometric derivatives are introduced later in this block.

Common Exam Pitfalls

• Substituting into the original expression before simplifying. If direct substitution gives $\frac{0}{0}$, that is not "the answer is zero" — it is a signal that algebraic work (factoring or rationalising) is required first.
• Treating $\infty$ as a number you can do arithmetic with. Expressions like $\infty - \infty$ or $\dfrac{\infty}{\infty}$ are indeterminate: they must be rewritten (usually by dividing by the highest power of $x$) before any conclusion can be drawn.
• Forgetting to check both one-sided limits when a function is defined piecewise or involves $|x|$, and claiming a limit exists when the two sides actually disagree.
• Confusing $f(a)$ with $\displaystyle\lim_{x\to a}f(x)$. The limit can exist and differ from $f(a)$, or exist when $f(a)$ is undefined altogether — the two questions are independent.
• Cancelling factors without stating the restriction. Writing $\dfrac{(x-3)(x+1)}{(x-3)(x+3)} = \dfrac{x+1}{x+3}$ is only valid for $x \neq 3$; this restriction is exactly why the limit calculation remains meaningful even though the original expression was undefined there.
• Misapplying the degree-comparison rule at infinity by comparing leading terms before dividing through, especially when the polynomials are not already in standard descending-power form.

Summary

• A sequence $(a_n)$ converges to $L$ if $a_n$ gets arbitrarily close to $L$ for sufficiently large $n$; otherwise it diverges.
• $\displaystyle\lim_{x\to a} f(x) = L$ describes the behaviour of $f$ approaching $a$, independent of $f(a)$ itself.
• The two-sided limit exists iff the left-hand and right-hand limits exist and are equal.
• Limits respect sums, products, quotients (denominator limit $\neq 0$) and powers of functions whose individual limits exist.
• The $\frac{0}{0}$ form is resolved by factor-and-cancel or rationalising, never by direct substitution.
• For rational functions as $x\to\infty$, divide through by the highest power of $x$ in the denominator; the result depends only on comparing the degrees of numerator and denominator.
• The squeeze theorem proves $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x} = 1$, a result used throughout the rest of this block.