Computing limits

Chapter 1.1 said what a limit is. This one is about getting the number.

There are only four moves, and the shape of the expression tells you which to reach for. Start by substituting — not because it always works, but because what it produces tells you what to do next.

Move 1: substitute

If $f$ is built from polynomials, roots, exponentials, sines and cosines by the usual arithmetic, and the substitution produces an ordinary number, then that number is the limit.

$$\lim_{x \to 3} \left(x^2 - 4x + 1\right) = 9 - 12 + 1 = -2$$

This is not laziness; it is a theorem. Those functions are continuous on their domains, and continuity is precisely the statement that the limit equals the value. Chapter 1.3 makes that formal.

Substitution can produce three kinds of result, and each is an instruction:

Move 2: factor and cancel

When substitution gives $\tfrac{0}{0}$ in a rational function, the numerator and denominator share a factor of $(x - a)$. Find it, cancel it, substitute again.

$$\lim_{x \to 3}\frac{x^2 - 9}{x - 3} = \lim_{x \to 3}\frac{(x-3)(x+3)}{x - 3} = \lim_{x \to 3}(x + 3) = 6$$

The cancellation is legal for the reason chapter 1.1 gave: the two expressions agree everywhere except at $x = a$, and the limit never looks at $x = a$.

Why the shared factor must be there. If substituting $a$ makes the numerator zero, then $a$ is a root of it, and a polynomial with root $a$ has $(x - a)$ as a factor. Same for the denominator. So the $\tfrac{0}{0}$ form guarantees the cancellation exists — you are never searching blindly.

Move 3: rationalise

A root in the expression blocks factoring. Multiply by the conjugate instead, which converts a difference of roots into a difference of squares.

$$\lim_{x \to 0}\frac{\sqrt{1+x} - 1}{x} = \lim_{x \to 0}\frac{\sqrt{1+x} - 1}{x}\cdot\frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} = \lim_{x \to 0}\frac{x}{x\left(\sqrt{1+x} + 1\right)} = \frac{1}{2}$$

The middle step is the whole trick: $(\sqrt{1+x}-1)(\sqrt{1+x}+1) = (1+x) - 1 = x$, and that $x$ cancels the one downstairs. The conjugate does for roots exactly what factoring does for polynomials — it produces the shared factor that the $\tfrac{0}{0}$ form promised.

Move 4: recognise

Some limits yield to no algebra at all. These are worth memorising, because the exam expects them and no manipulation produces them:

$$\lim_{x \to 0}\frac{\sin x}{x} = 1 \qquad \lim_{x \to 0}\frac{1 - \cos x}{x} = 0 \qquad \lim_{x \to 0}\frac{e^x - 1}{x} = 1$$

Note the fourth: $\tfrac{1-\cos x}{x} \to 0$ but $\tfrac{1-\cos x}{x^2} \to \tfrac{1}{2}$. The numerator vanishes like $x^2$, not like $x$ — so which denominator you divide by changes the answer entirely. Students who memorise only the first of those get the second wrong every time.

These three are the derivatives of $\sin$, $\cos$ and $e^x$ at zero, which is why they turn up everywhere in chapter 2.

Composing them

Most real limits need two moves. $\dfrac{\sin 5x}{x}$ is not a standard limit, but it becomes one after a rearrangement:

$$\frac{\sin 5x}{x} = 5 \cdot \frac{\sin 5x}{5x}$$

and as $x \to 0$ the inner argument $5x \to 0$ too, so the fraction tends to $1$ and the whole thing to $5$.

The pattern: make the argument and the denominator match, then the standard limit applies and the constant comes out front.

Practice