Examples of functions Continuous at a point

At every point of their domain, the following standard families of functions are continuous — each fact follows directly from the algebra-of-limits theorems (Section 9.2.3) applied at a general point $x_0$:

1. Constant functions. $f(x)=k$ is continuous at every $x_0\in\mathbb R$, since $\lim_{x\to x_0} k=k=f(x_0)$ trivially.
2. Power functions with positive-integer exponent. $f(x)=x^n$ is continuous everywhere, since $\lim_{x\to x_0} x^n=x_0^{\,n}$ by the algebra-of-limits power rule.
3. Polynomials. $p(x)=a_0+a_1x+\cdots+a_nx^n$ is continuous everywhere, being a finite sum of constant multiples of power functions.
4. Rational functions. $R(x)=\dfrac{p(x)}{q(x)}$ is continuous at every point where $q(x)\ne0$, by the quotient rule of the algebra of limits.
5. $\sin x$ and $\cos x$ are continuous on all of $\mathbb R$; consequently, by the quotient/reciprocal rules, $\tan x$, $\cot x$, $\sec x$, $\csc x$ are each continuous on their proper domains — wherever the relevant denominator (e.g. $\cos x$ for $\tan x$) is nonzero.
6. $n$th-root functions, $f(x)=x^{1/n}$, are continuous on their proper domain.
7. The reciprocal function $f(x)=1/x$ is undefined — hence not continuous — at $0$, but is continuous at every other point of $\mathbb R\setminus\{0\}$.
8. A genuinely piecewise example, $h(x)=x+1$ for $x\le0$ and $h(x)=x^2+1$ for $x>0$: checking directly at the join point, $\lim_{x\to0^-}h(x)=1=h(0)$ and $\lim_{x\to0^+}h(x)=1=h(0)$ as well, so $h$ is continuous even at $x=0$ (and obviously continuous elsewhere, being built from polynomials on each side) — continuous on the whole of $\mathbb R$.
9. The greatest integer function $f(x)=\lfloor x\rfloor$ is discontinuous at every integer $n$: $\lim_{x\to n^-}\lfloor x\rfloor=n-1$ while $\lim_{x\to n^+}\lfloor x\rfloor=n$ — the one-sided limits disagree at every integer point (matching the earlier discussion in Section 9.2.1) — a standing example of a function with infinitely many discontinuities, all of the "jump" type.
10. The modulus function $f(x)=|x|$, by contrast, is continuous everywhere on $\mathbb R$, including at $x=0$: $\lim_{x\to0^-}|x|=\lim_{x\to0^-}(-x)=0$ and $\lim_{x\to0^+}|x|=\lim_{x\to0^+}x=0$, both equal to $f(0)=0$ — so despite having a "corner" at the origin (which matters later for differentiability, but not for continuity), $|x|$ has no break in its graph there. …