continuous

Continuity — sequential definition

Let $D\subset \mathbb{R}$, $x_0\in D$ and $f:D\to \mathbb{R}$. We call the function $f$ continuous at $x_0$ if for all sequences $(x_n{)}_{n\in \mathbb{N}}$ with $x_n\to x_0$:

$$\underset{n\to \infty }{\lim }f(x_n)=f(x_0)=f\left(\underset{n\to \infty }{\lim }{x}_n\right)$$

If $f$ is continuous at all $x\in D$, we call $f$ continuous.

Continuous means: you can swap the limit and the function.

Continuity — Epsilon-delta criterion

$f$ is continuous at $x_0$ iff:

$$\forall \epsilon >0\exists \delta >0\forall x\in D:|x-x_0|<\delta ⟹|f(x)-f(x_0)|<\epsilon$$

“A small change in input causes only a small change in output.”

How to prove continuity

Using the sequential definition: take any sequence $x_n\to x_0$, apply limit rules to $f(x_n)$, show the result equals $f(x_0)$.

Using epsilon-delta: given $\epsilon >0$, find a $\delta >0$ (which may depend on $\epsilon$ and $x_0$) such that the implication holds.

How to prove discontinuity

Find one sequence $x_n\to x_0$ where $f(x_n)\nrightarrow f(x_0)$.

Or find two sequences $x_n,y_n\to x_0$ where $f(x_n)$ and $f(y_n)$ converge to different limits. This is stronger: it shows ${\lim }_{x\to x_0}f(x)$ doesn’t exist at all.

Removable vs non-removable discontinuity

Removable: ${\lim }_{x\to x_0}f(x)$ exists, but $\ne f(x_0)$ (or $f(x_0)$ is undefined). You can “fix” it by (re)defining $f(x_0):=$ the limit.
E.g. $\frac{x^2-4}{x-2}$ at $x=2$ is undefined, but ${\lim }_{x\to 2}={\lim }_{x\to 2}(x+2)=4$. Define $f(2)=4$ and it’s continuous.

Non-removable: ${\lim }_{x\to x_0}f(x)$ doesn’t exist. No choice of $f(x_0)$ can fix it. This happens when:

• left and right limits disagree (jump, e.g. heaviside function at 0)
• the function oscillates without settling (e.g. $\sin (1/x)$ at 0)

Calculation rules

If $f,g:D\to \mathbb{R}$ are continuous at $x_0$ and $c\in \mathbb{R}$, then so are:

• $f+g$, $f\cdot g$, $c\cdot f$
• $\frac{f}{g}$ (if $g(x_0)\ne 0$)
• $g\circ f$ (composition, if the domains match)

Follows directly from limit arithmetic.

Known continuous functions

These are the building blocks. Anything composed from them via the calculation rules above is also continuous.

• Constants, $x\mapsto x$ (identity)
• polynomials (by applying calculation rules to $x$ and constants)
• Rational functions $p(x)/q(x)$ (where $q(x)\ne 0$)
• $\sin x$, $\cos x$ (and therefore $\tan x$ where $\cos x\ne 0$)
• $a^x$ (exponential, for $a>0$)
• ${\log }_{b}x$ (for $x>0$, $b>1$)
• $\sqrt[k]{x}$ (for $x\ge 0$)
• $|x|$

Continuity of piecewise functions

Each piece is typically continuous on its interval (polynomial, sqrt, sin, …). The only question is at the junction points where the pieces meet. There, check:

$$\underset{x\to x_0^{-}}{\lim }f(x)=\underset{x\to x_0^{+}}{\lim }f(x)=f(x_0)$$

Left limit = right limit = function value. If any of these disagree, $f$ is discontinuous at $x_0$.

Inverse functions

If $f:I\to D$ is continuous and bijective on an interval $I$, then $f^{-1}$ is also continuous on $D$. This is why $\sqrt[k]{x}$ and ${\log }_{b}x$ are continuous (they’re inverses of $x^k$ and $b^x$).

Continuity $\ne$ Differentiability

If a function is differentiable at a point, then it must be continuous at that point.
However, a function can be continuous at a point without being differentiable there.
E.g. $f(x)=|x|$ is continuous everywhere but not differentiable at $x=0$; $f(x)=x^{1/3}$ has a vertical tangent at $x=0$, i.e. not differentiable there.

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