Table of Contents

Single Variable Calculus — Lecture 1

Opening lecture notes for a one-variable calculus course: limits, continuity, and the first theorems you actually prove from the $\varepsilon$-$\delta$ definition.

Limits

Let $f$ be a function defined on an open interval containing $a$, except possibly at $a$ itself. We say that

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

if for every $\varepsilon > 0$ there exists $\delta > 0$ such that whenever $0 < |x - a| < \delta$, we have $|f(x) - L| < \varepsilon$.

Intuitively, $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $a$.

Limit of a sum. If $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$, then

$$\lim_{x \to a} \bigl(f(x) + g(x)\bigr) = L + M.$$

Let $\varepsilon > 0$. Choose $\delta_1 > 0$ so that $|f(x) - L| < \varepsilon/2$ when $0 < |x - a| < \delta_1$, and $\delta_2 > 0$ so that $|g(x) - M| < \varepsilon/2$ when $0 < |x - a| < \delta_2$.

Take $\delta = \min(\delta_1, \delta_2)$. Then for $0 < |x - a| < \delta$,

$$|(f(x) + g(x)) - (L + M)| \le |f(x) - L| + |g(x) - M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

Continuity

A function $f$ is continuous at $a$ if

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

If $f$ is differentiable at $a$, then $f$ is continuous at $a$.

Assume $f$ is differentiable at $a$. For $x \ne a$, write

$$f(x) - f(a) = \frac{f(x) - f(a)}{x - a}(x - a).$$

Taking the limit as $x \to a$ and using $f'(a) = \displaystyle\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ gives $\displaystyle\lim_{x \to a} f(x) = f(a)$.

The squeeze theorem

Suppose $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$). If

$$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L,$$

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

Given $\varepsilon > 0$, pick $\delta > 0$ so that $L - \varepsilon < g(x) < L + \varepsilon$ and $L - \varepsilon < h(x) < L + \varepsilon$ whenever $0 < |x - a| < \delta$. Then

$$L - \varepsilon < g(x) \le f(x) \le h(x) < L + \varepsilon,$$

so $|f(x) - L| < \varepsilon$.

A classic limit

$$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$

Geometrically, compare the areas of a sector, a triangle, and a larger triangle in the unit circle; the squeeze theorem finishes the argument. (Left as an exercise — draw the picture.)

Derivative preview

The derivative at $a$ is the linear approximation rate:

$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}.$$

Next lecture: rules of differentiation and the chain rule.