Limits

A limit describes what a function approaches, not what it equals. That distinction is why $\dfrac{x^2-4}{x-2}$ has a perfectly good limit at $x = 2$ even though the function is undefined there — and it is the idea the whole of calculus is built on.

The idea of a limit

$$\lim_{x \to a} f(x) = L$$

means: as $x$ gets arbitrarily close to $a$ (from either side), $f(x)$ gets arbitrarily close to $L$.

The value $f(a)$ is irrelevant to the limit. The function may be undefined at $a$, or defined with a completely different value, and the limit is unaffected.

For $f(x) = \dfrac{x^2-4}{x-2}$, substituting $x = 2$ gives $\frac00$ — meaningless. But for every $x \ne 2$,

$$\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2$$

so as $x \to 2$ the function approaches $4$.

$$\lim_{x\to2}\frac{x^2-4}{x-2} = 4$$

One-sided limits

$\lim_{x\to a^-}$ approaches from the left, $\lim_{x\to a^+}$ from the right. The two-sided limit exists only if both agree.

For $f(x) = \dfrac{|x|}{x}$, the left limit at 0 is $-1$ and the right limit is $+1$. They differ, so $\lim_{x\to0} f(x)$ does not exist.

Limit laws

Provided the individual limits exist:

$$\lim(f \pm g) = \lim f \pm \lim g$$ $$\lim(fg) = (\lim f)(\lim g)$$ $$\lim\frac{f}{g} = \frac{\lim f}{\lim g} \quad \text{provided } \lim g \ne 0$$ $$\lim(kf) = k\lim f$$

For a polynomial, or any function continuous at $a$, the limit is simply the value:

$$\lim_{x\to2}(3x+1) = 7$$

Evaluating limits

Step 1: substitute. If you get a number, that is the answer.

Step 2: if you get $\frac00$ — an indeterminate form — the expression must be simplified before substituting.

By factorising:

$$\lim_{x\to3}\frac{x^2-9}{x^2-x-6} = \lim_{x\to3}\frac{(x-3)(x+3)}{(x-3)(x+2)} = \lim_{x\to3}\frac{x+3}{x+2} = \frac65$$

By rationalising, when roots are involved:

$$\lim_{x\to0}\frac{\sqrt{x+4}-2}{x} = \lim_{x\to0}\frac{(x+4)-4}{x\left(\sqrt{x+4}+2\right)} = \lim_{x\to0}\frac{1}{\sqrt{x+4}+2} = \frac14$$

Limits at infinity

Divide every term by the highest power of $x$ in the denominator, then use $\dfrac{1}{x^n} \to 0$.

$$\lim_{x\to\infty}\frac{3x^2+2x}{x^2-5} = \lim_{x\to\infty}\frac{3 + \frac2x}{1 - \frac{5}{x^2}} = \frac{3}{1} = 3$$

$$\lim_{x\to\infty}\frac{2x+1}{x^2+3} = \lim_{x\to\infty}\frac{\frac2x + \frac{1}{x^2}}{1 + \frac{3}{x^2}} = 0$$

The pattern for rational functions:

Special trigonometric limits

$$\lim_{\theta\to0}\frac{\sin\theta}{\theta} = 1 \qquad \lim_{\theta\to0}\frac{1-\cos\theta}{\theta} = 0$$

with $\theta$ in radians. The first says that for small angles $\sin\theta \approx \theta$ — which is why radians are the natural unit for calculus.

Adapting the first result:

$$\lim_{x\to0}\frac{\sin 3x}{x} = \lim_{x\to0}\ 3 \cdot \frac{\sin 3x}{3x} = 3(1) = 3$$

$$\lim_{x\to0}\frac{\sin 5x}{\sin 2x} = \lim_{x\to0}\frac{\frac{\sin 5x}{5x}\cdot 5x}{\frac{\sin 2x}{2x}\cdot 2x} = \frac{5}{2}$$

Continuity

$f$ is continuous at $x = a$ if all three of these hold:

1. $f(a)$ is defined;
2. $\lim_{x\to a} f(x)$ exists;
3. $\lim_{x\to a} f(x) = f(a)$.

Informally: you can draw the graph through that point without lifting your pen.

$\dfrac{x^2-4}{x-2}$ is not continuous at $x = 2$ — the limit exists but $f(2)$ does not. This is a removable discontinuity: defining $f(2) = 4$ repairs it.

Polynomials are continuous everywhere. Rational functions are continuous except where the denominator vanishes.

The derivative from first principles

Every derivative rule you will meet comes from one limit:

$$f'(x) = \lim_{h\to0}\frac{f(x+h) - f(x)}{h}$$

This is the gradient of a chord as its width shrinks to zero.

Differentiate $f(x) = x^2$ from first principles.

$$f'(x) = \lim_{h\to0}\frac{(x+h)^2 - x^2}{h}$$ $$= \lim_{h\to0}\frac{x^2 + 2xh + h^2 - x^2}{h}$$ $$= \lim_{h\to0}\frac{2xh + h^2}{h}$$ $$= \lim_{h\to0}(2x + h)$$ $$= 2x$$

And $f(x) = x^3$:

$$\frac{(x+h)^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2 \ \longrightarrow \ 3x^2$$