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Let $f$ be differentiable on an open interval containing $c$. If the finite limit of a function $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, then $f^{\prime }(c)=L$.

Choose $h_n>0$ tending to zero with $[c,c+h_n]$ in the interval. The mean value theorem gives $x_n\in (c,c+h_n)$ such thatThe right-hand side tends to $f^{\prime }(c)$ by differentiability. Since $x_n\to c$ with $x_n\ne c$, the left-hand side tends to $L$. Uniqueness of the limit of a sequence gives equality. The same reasoning excludes an infinite limit of a function for $f^{\prime }$ at $c$.

Thus derivatives may be discontinuous, but a removable discontinuity cannot be repaired by changing the value of a derivative at a single point of differentiability.

A function $f:D\to \mathbb{R}$ is continuous at $a\in D$ whenFirst choose the required output tolerance $\epsilon$; then choose an input tolerance $\delta$ that works for every allowed function input in that window. Picking a different $\delta$ separately after seeing each $x$ would reverse the quantifiers and lose this guarantee. There is no need to exclude $x=a$, because the output difference is then zero.

At an isolated point of the domain of a function, choose a small enough window that its only allowed function input is $a$. This proves continuity there directly. At a point with arbitrarily close other inputs, the same definition is the limit of a function condition $f(x)\to f(a)$. For metric spaces, replace the two absolute values by their distance functions; for arbitrary topological spaces, the corresponding definition is a continuous map.

The input tolerance may depend on $a$. A single choice valid everywhere on the domain of a function is the stronger requirement of uniform continuity. The continuous function $x\mapsto x^2$ on $\mathbb{R}$ shows the difference: for any proposed $\delta >0$, the points $a=1/\delta$ and $b=a+\delta /2$ have $|b-a|<\delta$, yet $b^2-a^2=1+{\delta }^2/4>1$. Thus the fixed output tolerance $\epsilon =1$ admits no global input tolerance on the real line.

If a function is differentiable at an interior local minimum or local maximum, its derivative there is zero. At a local minimum $c$, for small $h>0$ the difference quotient $(f(c+h)-f(c))/h$ is nonnegative, while for small $h<0$ it is nonpositive. The one-sided limit of a function from each side equals $f^{\prime }(c)$, forcing $f^{\prime }(c)\ge 0$ and $f^{\prime }(c)\le 0$. Apply the same argument to $-f$ at a local maximum.

The converse fails: $f(x)=x^3$ has $f^{\prime }(0)=0$ but no local minimum or local maximum at zero. The interior hypothesis is essential; a minimum attained only at an endpoint need not have derivative zero.

For $n=1$, differentiability at $a$ means that the difference quotient has a finite two-sided limit of a function. For $n\ge 2$, require derivatives through order $n-1$ throughout some open neighbourhood of $a$, and require the final difference quotientto exist as a finite real number. The last derivative need not exist elsewhere or be continuous at $a$. In particular, existence of $f^{\prime }(a)$ by itself does not justify taking a second derivative: the function $f^{\prime }$ must first be defined in a neighbourhood of $a$.

For $n=1$, differentiability at $x_0$ means existence of the finite two-sided limit of a function ${\lim }_{h\to 0}(f(x_0+h)-f(x_0))/h$. More generally, higher-order differentiability at a point is defined recursively: for $n\ge 2$, the derivatives $f^{\prime },\dots ,f^{(n-1)}$ must exist throughout some open interval containing $x_0$, and $f^{(n-1)}$ must be differentiable at $x_0$. With $f^{(0)}=f$, the final condition isThe neighbourhood requirement makes the numerator meaningful for every sufficiently small nonzero $h$. It does not require $f^{(n)}$ to exist at other points of that neighbourhood, or to be continuous at $x_0$.

For $0<x<\pi /2$, the usual unit-circle area comparison, with angles measured in radians, gives $\sin x\le x\le \tan x$. HenceThe cosine is a continuous function at zero, so the squeeze theorem yields the right-hand limit of a function $1$. The quotient $\sin x/x$ is even, so the left-hand limit of a function is the same. Thus the limit is $1$.

Let $A_n$ be the event that the origin has an open graph path to the boundary of $[-n,n{]}^d\cap {\mathbb{Z}}^d$. The graph path can be stopped on its first boundary visit, so $A_n$ depends on only finitely many sites. Its probability $f_n(p)={\mathbb{P}}_p(A_n)$ is a finite sum of terms $p^a(1-p{)}^b$, and is therefore a continuous function of $p$. Also $A_{n+1}\subseteq A_n$, and the König infinity lemma gives ${\bigcap }_n{A}_n=\{0\leftrightarrow \infty \}$. By continuity from above of a measure,The monotone coupling of Bernoulli percolation described below shows that $\theta$ is nondecreasing. For $p<1$, its right limit of a function $\theta (p+)$ exists and is at least $\theta (p)$. For every fixed $n$,Taking the infimum in $n$ proves $\theta (p+)=\theta (p)$. This proves right continuity of percolation probability throughout $[0,1)$; at $1$ right continuity is understood relative to the parameter domain. The argument does not interchange two uncontrolled limits.

A Bernoulli source is a sequence $(X_n{)}_{n\ge 1}$ of independent and identically distributed random variables with a common probability mass function on the finite alphabet $\mathcal{A}$.

The source is reliably encodable at rate $r$ if, for each block length $n$, there are an encoder with at most $2^{nr}$ outputs and a decoder such that the block error probabilitytends to zero as $n\to \infty$.

Its information rate is the limiting information entropy per source symbol,when this limit exists.

On $(0,1)$, the product rule and chain rule give $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. On $(-1,0)$, the function is constant and has derivative zero. At zero,Its absolute value is at most $|h|$, so the squeeze theorem gives $f^{\prime }(0)=0$. ThereforeFor $x_n=1/(2\pi n)$, we have $x_n\to 0$ but $f^{\prime }(x_n)=-1$ for every $n$. This differs from $f^{\prime }(0)=0$, so $f^{\prime }$ is not continuous at zero. Indeed, $y_n=1/((2n+1)\pi )$ gives $f^{\prime }(y_n)=1$, so the right-hand limit of a function does not exist. Thus $f$ is differentiable everywhere on $(-1,1)$, while $f^{\prime }$ is discontinuous at zero.

A function $f:(a,b)\to \mathbb{R}$ is differentiable on $(a,b)$ when, at each $c\in (a,b)$, its difference quotient has a finite limit of a function:Here $h\ne 0$ and $c+h\in (a,b)$.

For sufficiently small nonzero $h$, the difference quotient has absolute value at most $|f^{\prime }(c)|+1$, because it tends to $f^{\prime }(c)$. ConsequentlyThis is continuity at $c$. Since $c$ was arbitrary, differentiability implies continuity on $(a,b)$.

Choose $h_n=\min \{1/n,(b-c)/2\}>0$. Then $h_n\to 0$ and $[c,c+h_n]\subset (a,b)$. The mean value theorem states that a continuous function on a closed interval, differentiable in its interior, has an interior derivative equal to its endpoint difference quotient. It gives $x_n\in (c,c+h_n)$ withHere $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f^{\prime }(c)$ by differentiability at $c$. If $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, the same values $f^{\prime }(x_n)$ must tend to $L$. Uniqueness of the limit of a sequence givesThus a derivative cannot have a removable discontinuity at a point where the original function is differentiable. The proof also excludes an infinite limit of a function there, since the displayed difference quotients have a finite limit of a sequence.

The function $f$ is differentiable at $c$ if the finite limit of a functionexists. If it exists, put $L=f^{\prime }(c)$ and defineThe definition of derivative gives $ϵ(x)\to 0$ as $x\to c$. For $x\ne c$, rearrangement gives the desired local linear approximationAt $x=c$, this identity holds because both terms containing $x-c$ vanish. Conversely, if such an identity holds, then for $x\ne c$ its difference quotient is $L+ϵ(x)\to L$. Hence $f$ is differentiable at $c$, and necessarily $\boxed{L=f^{\prime }(c)}$.

A function $f:(a,b)\to \mathbb{R}$ is differentiable on $(a,b)$ if, for every $x\in (a,b)$, the finite limit of a functionexists. Here $h\ne 0$ and is sufficiently small that $x+h\in (a,b)$. This limit of a function is the derivative of $f$ at $x$.

We give the explicit point $\boxed{x=1/4}$. Repeatedly choose the left remaining third and then the right remaining third. After $2m$ choices, the resulting closed interval is $[L_m,R_m]$, whereThese formulas follow from a finite geometric series. They show $L_m<1/4<R_m$, and that both endpoints tend to $1/4$. The point $1/4$ is also strictly inside each intermediate left third: its distance from $L_m$ is one quarter of the current length. It therefore never enters any assigned closed middle third or its endpoints, so $1/4\in C$.

Endpoint values in the Cantor function construction obey the same left/right choices, with halving in place of thirding. ThusAll these endpoints are in the domain of the partial function: $L_m$ is the right endpoint of a middle third assigned at round $2m$, and $R_m$ is the left endpoint of a middle third assigned at round $2m-1$.

Given $\epsilon >0$, choose $m$ so that $(2/3)4^{-m}<\epsilon$, and put $\delta =9^{-m}/4$. If $t\notin C$ and $|t-1/4|<\delta$, then $L_m<t<R_m$. The assumed monotonicity givesConsequently $|f(t)-1/3|<\epsilon$. This proves directly, on the stated partial domain,There are defined points arbitrarily close to $1/4$, for example $L_m$ and $R_m$, so the limit of a function is not vacuous. No continuous extension has been assumed. The residual $C$ here omits the endpoints that the question assigns values to; it is the usual Cantor set with those assigned endpoints removed.

The limit of a function $f:E\to \mathbb{R}$ at a limit point $p$ of $E$ is $\ell$ whenThe function converges as $x\to p$ if such a finite $\ell$ exists. The puncture excludes any value assigned at $p$.

For $p\in E$, continuity at $p$ meansWhen $p$ is also a limit point, this says that the limit of a function is $f(p)$. At an isolated point of $E$, every function is continuous relative to $E$, even though the punctured limit does not have a unique forced value.

Continuity at $x_0$ requires that $f(x_0)$ be defined and that ${\lim }_{x\to x_0}f(x)=f(x_0)$. In the epsilon-delta definition, for each $\epsilon >0$ there must be a $\delta >0$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$ within the domain. Differentiability requires the finite two-sided limitIn particular, merely having a punctured limit of a function does not imply continuity unless the assigned value agrees with that limit.

Continuity. The function is continuous at $x_0$ when ${\lim }_{x\to x_0}f(x)=f(x_0)$. Equivalently, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ within the domain of a function implies $|f(x)-f(x_0)|<\epsilon$; this is the epsilon-delta definition of continuity.

Differentiability. At an interior point $x_0$, the difference quotient must have a finite two-sided limit of a function:This defines differentiability and the derivative. At an endpoint one may instead state a one-sided definition relative to the interval. Differentiability implies continuity, because the increment is $h$ times a quotient with finite limit, but continuity alone does not guarantee differentiability.