Limits
Concept and Algebra of Limits
A limit describes the value a function approaches as its input creeps towards some number $a$ — not the value at $a$, which the function may never actually reach (or may not even be defined there). We write $\lim_{x \to a} f(x) = L$ to mean: as $x$ gets arbitrarily close to $a$, the output $f(x)$ gets arbitrarily close to $L$.
One-sided limits. Because $x$ can approach $a$ from two directions, we distinguish the left-hand limit (LHL), approaching through values smaller than $a$, and the right-hand limit (RHL), approaching through larger values:
Existence of a limit. The limit at $a$ exists if and only if both one-sided limits exist and agree. This is the single most-tested idea of the topic:
If the two sides disagree — as happens at a jump — the limit simply does not exist, no matter how nicely the function behaves elsewhere.
Algebra of limits. Provided $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist, limits respect the basic operations. Let $\lim f = L$ and $\lim g = M$:
| Law | Statement |
|---|---|
| Sum / Difference | $\lim (f \pm g) = L \pm M$ |
| Constant multiple | $\lim (k\,f) = kL$ |
| Product | $\lim (f \cdot g) = L \cdot M$ |
| Quotient | $\lim \dfrac{f}{g} = \dfrac{L}{M}, \;\; M \ne 0$ |
| Power | $\lim \big(f\big)^{n} = L^{n}$ |
Polynomials and rational functions. Polynomials are continuous everywhere, so for any polynomial $p(x)$ you get the limit by direct substitution: $\lim_{x \to a} p(x) = p(a)$. For a rational function $\dfrac{p(x)}{q(x)}$, substitution works as long as $q(a) \ne 0$.
The $\dfrac{0}{0}$ form. When both numerator and denominator vanish at $a$, substitution gives the meaningless symbol $\dfrac{0}{0}$ — an indeterminate form. The cure is to factorise and cancel the common factor $(x - a)$ that is causing both to vanish, then substitute into what remains:
Deeper Insight — a limit is about the journey, not the destination: The whole power of the limit concept comes from a deliberate separation between what a function does near a point and what it is at that point. In the example above, $\dfrac{x^2-4}{x-2}$ is genuinely undefined at $x = 2$ — there is a hole in the graph — yet the limit is a perfectly definite $4$, because every nearby value of the function is close to $4$. This is exactly why cancelling the factor $(x-2)$ is legitimate: for all $x \ne 2$ the two expressions are identical, and the limit only ever cares about $x \ne a$. The $\dfrac{0}{0}$ symbol is not an answer but a signal that a hidden common factor is masking the true behaviour, and learning to read that signal — factor, rationalise, or simplify — is the skill that the entire calculus syllabus is built upon.
- A polynomial is continuous everywhere, so use direct substitution.
- $2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4$.
Answer: $\lim_{x \to 3}\,(2x^2 - 5x + 1) = 4$.
- Substituting $x = 1$ gives $\dfrac{0}{0}$ — an indeterminate form.
- Factorise the numerator: $x^2 - 1 = (x - 1)(x + 1)$.
- Cancel the common factor $(x - 1)$: $\dfrac{(x-1)(x+1)}{x-1} = x + 1$ for $x \ne 1$.
- Now substitute: $\lim_{x \to 1}(x + 1) = 2$.
Answer: $\lim_{x \to 1}\dfrac{x^2 - 1}{x - 1} = 2$.
- Left-hand limit: $\lim_{x \to 0^-}(x + 2) = 0 + 2 = 2$.
- Right-hand limit: $\lim_{x \to 0^+}(3 - x) = 3 - 0 = 3$.
- Since LHL $= 2 \ne 3 = $ RHL, the one-sided limits disagree.
Answer: The limit does not exist at $x = 0$ (a jump).
- Substituting $x = 2$ gives $\dfrac{0}{0}$, so factorise both parts.
- Numerator: $x^2 - 5x + 6 = (x - 2)(x - 3)$.
- Denominator: $x^2 - 4 = (x - 2)(x + 2)$.
- Cancel $(x - 2)$: $\dfrac{x - 3}{x + 2}$, then substitute: $\dfrac{2 - 3}{2 + 2} = -\dfrac{1}{4}$.
Answer: $\lim_{x \to 2}\dfrac{x^2 - 5x + 6}{x^2 - 4} = -\dfrac{1}{4}$.
- Direct substitution gives $\dfrac{0}{0}$; rationalise by multiplying by the conjugate $\sqrt{1+x}+1$.
- $\dfrac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \dfrac{(1+x) - 1}{x(\sqrt{1+x}+1)} = \dfrac{x}{x(\sqrt{1+x}+1)}$.
- Cancel $x$: $\dfrac{1}{\sqrt{1+x}+1}$, then substitute $x = 0$: $\dfrac{1}{1 + 1} = \dfrac{1}{2}$.
Answer: $\lim_{x \to 0}\dfrac{\sqrt{1 + x} - 1}{x} = \dfrac{1}{2}$.
- Apply the algebra of limits term by term.
- Numerator: $3(4) + (-2) = 12 - 2 = 10$.
- Denominator: $f \cdot g \to (4)(-2) = -8$, which is non-zero, so the quotient law applies.
- Divide: $\dfrac{10}{-8} = -\dfrac{5}{4}$.
Answer: $\lim_{x \to a}\dfrac{3f(x) + g(x)}{f(x)\,g(x)} = -\dfrac{5}{4}$.
- A limit captures what $f(x)$ approaches near $a$, not its value at $a$.
- $\lim_{x \to a} f(x)$ exists only if LHL $=$ RHL.
- For polynomials and well-defined rational functions, use direct substitution.
- The $\dfrac{0}{0}$ form is indeterminate — factorise and cancel (or rationalise) the common factor.
- Limits distribute over sums, products and quotients provided each piece exists (denominator $\ne 0$).
Standard Limits
Some limits arise so often that they are worth committing to memory as standard results. Each is itself a $\dfrac{0}{0}$ form that cannot be cracked by simple cancellation, yet each settles to a clean value. Knowing them turns long derivations into one-line substitutions.
The trigonometric limit. The cornerstone of calculus with angles — valid only when $x$ is measured in radians:
An immediate companion follows from it: $\lim_{x \to 0}\dfrac{\tan x}{x} = 1$, since $\tan x = \dfrac{\sin x}{\cos x}$ and $\cos x \to 1$.
The exponential and logarithmic limits. These power the derivatives of $e^x$, $a^x$ and $\ln x$:
The power limit. A purely algebraic result that generalises difference-of-powers factorisation to any real index $n$:
Collected together for quick reference:
| Standard limit | Value |
|---|---|
| $\lim_{x \to 0}\dfrac{\sin x}{x}$ | $1$ |
| $\lim_{x \to 0}\dfrac{\tan x}{x}$ | $1$ |
| $\lim_{x \to 0}\dfrac{1 - \cos x}{x^2}$ | $\dfrac{1}{2}$ |
| $\lim_{x \to 0}\dfrac{e^x - 1}{x}$ | $1$ |
| $\lim_{x \to 0}\dfrac{a^x - 1}{x}$ | $\ln a$ |
| $\lim_{x \to 0}\dfrac{\log(1 + x)}{x}$ | $1$ |
| $\lim_{x \to a}\dfrac{x^n - a^n}{x - a}$ | $n\,a^{\,n-1}$ |
How to use them. The trick is to massage the expression until it matches a standard shape exactly. To evaluate $\lim_{x \to 0}\dfrac{\sin 5x}{x}$, write it as $5 \cdot \dfrac{\sin 5x}{5x}$; as $x \to 0$ the angle $5x \to 0$ too, so $\dfrac{\sin 5x}{5x} \to 1$ and the limit is $5$.
Deeper Insight — why radians, and why these limits are the seeds of differentiation: The result $\lim_{x \to 0}\tfrac{\sin x}{x}=1$ holds only in radians, and that is no accident: the radian is defined precisely so that, for tiny angles, the arc length and the vertical sine are practically equal, which is what makes the ratio tend to $1$. Switch to degrees and you would instead get $\tfrac{\pi}{180}$, contaminating every trigonometric derivative — this is the deep reason calculus always uses radians. More broadly, every standard limit here is really a derivative in disguise evaluated at a single point: $\lim_{x\to 0}\tfrac{e^x-1}{x}$ is the slope of $e^x$ at $x=0$, and $\lim_{x\to a}\tfrac{x^n-a^n}{x-a}$ is the slope of $x^n$ at $x=a$. Recognising them as building blocks, rather than isolated facts to memorise, is what lets you derive the rules of the next topic instead of merely recalling them.
- Reshape to match $\dfrac{\sin(\cdot)}{(\cdot)}$: multiply and divide by $5$.
- $\dfrac{\sin 5x}{x} = 5 \cdot \dfrac{\sin 5x}{5x}$.
- As $x \to 0$, the angle $5x \to 0$, so $\dfrac{\sin 5x}{5x} \to 1$.
- Therefore the limit is $5 \times 1 = 5$.
Answer: $\lim_{x \to 0}\dfrac{\sin 5x}{x} = 5$.
- Force each sine over its own angle by inserting $3x$ and $7x$.
- $\dfrac{\sin 3x}{\sin 7x} = \dfrac{\sin 3x}{3x} \cdot \dfrac{7x}{\sin 7x} \cdot \dfrac{3x}{7x}$.
- As $x \to 0$: $\dfrac{\sin 3x}{3x} \to 1$ and $\dfrac{7x}{\sin 7x} \to 1$, leaving $\dfrac{3}{7}$.
Answer: $\lim_{x \to 0}\dfrac{\sin 3x}{\sin 7x} = \dfrac{3}{7}$.
- Use the identity $1 - \cos x = 2\sin^2\dfrac{x}{2}$.
- $\dfrac{2\sin^2(x/2)}{x^2} = \dfrac{2\sin^2(x/2)}{4\,(x/2)^2} = \dfrac{1}{2}\left(\dfrac{\sin(x/2)}{x/2}\right)^2$.
- As $x \to 0$, $\dfrac{\sin(x/2)}{x/2} \to 1$, so the bracket squared $\to 1$.
- Limit $= \dfrac{1}{2} \times 1 = \dfrac{1}{2}$.
Answer: $\lim_{x \to 0}\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}$.
- Reshape to match $\dfrac{e^{(\cdot)} - 1}{(\cdot)}$ by inserting $2x$.
- $\dfrac{e^{2x} - 1}{x} = 2 \cdot \dfrac{e^{2x} - 1}{2x}$.
- As $x \to 0$, $2x \to 0$, so $\dfrac{e^{2x} - 1}{2x} \to 1$.
- Limit $= 2 \times 1 = 2$.
Answer: $\lim_{x \to 0}\dfrac{e^{2x} - 1}{x} = 2$.
- This is the standard form $\lim_{x \to a}\dfrac{x^n - a^n}{x - a} = n\,a^{\,n-1}$ with $a = 2$ (since $32 = 2^5$).
- Here $n = 5$ and $a = 2$.
- Apply the formula: $n\,a^{\,n-1} = 5 \cdot 2^{4} = 5 \times 16 = 80$.
Answer: $\lim_{x \to 2}\dfrac{x^5 - 32}{x - 2} = 80$.
- This is the standard exponential form $\lim_{x \to 0}\dfrac{a^x - 1}{x} = \ln a$.
- Here $a = 3$.
- So the limit equals $\ln 3$.
Answer: $\lim_{x \to 0}\dfrac{3^x - 1}{x} = \ln 3$.
- $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$ holds only in radians; $\lim_{x \to 0}\dfrac{\tan x}{x} = 1$ follows.
- $\lim_{x \to 0}\dfrac{e^x - 1}{x} = 1$ and $\lim_{x \to 0}\dfrac{a^x - 1}{x} = \ln a$.
- $\lim_{x \to a}\dfrac{x^n - a^n}{x - a} = n\,a^{\,n-1}$ for any real index $n$.
- To use a standard limit, massage the expression so the angle (or exponent) and its denominator match exactly.
- $\lim_{x \to 0}\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}$ is a frequent supporting result.
Infinity and Indeterminate Forms
For limits at infinity of rational functions, compare degrees (divide by the highest power). Forms like 0/0 and ∞/∞ need algebra first.
- At infinity, compare degrees of numerator and denominator.
- Resolve 0/0 by factor/rationalise before substituting.