After pointwise continuity, the next step is to separate limiting behaviour from the actual value of a function. Limits of mappings describe what $f(x)$ approaches as $x$ approaches a point $a$, even when $f(a)$ is undefined or irrelevant. The focus keyword limits of mappings is essential here because the domain may be a subspace and the point of approach may lie on its boundary. Continuity at a limit point is then the special case where the limit exists and equals the value of the function. :::definition[Limit of a Mapping] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $A\subseteq X$, let $f:A\to Y$, and let $a$ be a limit point of $A$. A point $b\in Y$ is called the limit of $f(x)$ as $x$ tends to $a$ if for every $\epsilon>0$, there exists $\delta>0$ such that $$\begin{equation*} d_Y(f(x),b)<\epsilon \end{equation*}$$ whenever $x\in A$ and $$\begin{equation*} 0<d_X(x,a)<\delta. \end{equation*}$$ In this case, we write $\lim_{x\to a}f(x)=b$. ::: :::remark The point $a$ need not belong to $A$. It is enough that $a$ is a limit point of $A$. If $a\in A$, the value $f(a)$ need not equal the limit of $f(x)$ as $x\to a$. ::: :::theorem[Sequential Criterion for Limits] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $A\subseteq X$, let $f:A\to Y$, and let $a$ be a limit point of $A$. Then $\lim_{x\to a}f(x)=b$ if and only if for every sequence $\{x_n\}$ in $A$ satisfying $x_n\neq a$ and $x_n\to a$, we have $f(x_n)\to b$. ::: :::proof Given that $a$ is a limit point of $A$. To prove that the epsilon-delta definition of the limit is equivalent to the sequential definition. First suppose that $\lim_{x\to a}f(x)=b$. Let $\{x_n\}$ be a sequence in $A$ such that $x_n\neq a$ and $x_n\to a$. Let $\epsilon>0$. There exists $\delta>0$ such that $d_Y(f(x),b)<\epsilon$ whenever $0<d_X(x,a)<\delta$. Since $x_n\to a$, there exists $n_0\in\mathbb N$ such that $d_X(x_n,a)<\delta$ for all $n\geq n_0$. Since $x_n\neq a$, we get $0<d_X(x_n,a)<\delta$. Therefore $d_Y(f(x_n),b)<\epsilon$ for all $n\geq n_0$. Hence, $f(x_n)\to b$. Conversely, suppose the sequential condition holds. To prove the limit condition, if possible let $\lim_{x\to a}f(x)\neq b$. Then there exists $\epsilon_0>0$ such that for every $\delta>0$, there exists $x\in A$ satisfying $0<d_X(x,a)<\delta$ and $d_Y(f(x),b)\geq\epsilon_0$. Taking $\delta=1/n$, choose $x_n\in A$ such that $0<d_X(x_n,a)<1/n$ and $d_Y(f(x_n),b)\geq\epsilon_0$. Then $x_n\neq a$ and $x_n\to a$, but $f(x_n)$ does not converge to $b$. A contradiction. Hence, $\lim_{x\to a}f(x)=b$. $\square$ ::: :::theorem[Continuity and Limits] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $A\subseteq X$, let $f:A\to Y$, and let $a\in A$ be a limit point of $A$. Then $f$ is continuous at $a$ if and only if $$\begin{equation*} \lim_{x\to a}f(x)=f(a). \end{equation*}$$ ::: :::proof Given that $a\in A$ is a limit point of $A$. To prove that continuity at $a$ is equivalent to equality of the limit with the value of the function. If $f$ is continuous at $a$, then the continuity inequality holds whenever $x\in A$ and $d_X(x,a)<\delta$. Therefore it also holds when $0<d_X(x,a)<\delta$, and $\lim_{x\to a}f(x)=f(a)$. Conversely, suppose $\lim_{x\to a}f(x)=f(a)$. Let $\epsilon>0$. There exists $\delta>0$ such that $d_Y(f(x),f(a))<\epsilon$ whenever $0<d_X(x,a)<\delta$. If $x=a$, then $d_Y(f(x),f(a))=0<\epsilon$. Therefore the continuity condition holds at $a$. Hence, $f$ is continuous at $a$. $\square$ ::: :::example[Limit Without Defined Value] Let $A=(0,1]$ and $f:A\to\mathbb R$ be defined by $f(x)=x^2$. The point $0$ is a limit point of $A$, although $0\notin A$. Given $\epsilon>0$, choose $\delta=\sqrt{\epsilon}$. If $0<|x|<\delta$, then $|x^2|<\epsilon$. Hence, $\lim_{x\to0}f(x)=0$. ::: :::solved-problem Let $f:\mathbb R\setminus\{0\}\to\mathbb R$ be defined by $f(x)=\sin(1/x)$. Show that $\lim_{x\to0}f(x)$ does not exist. ::: :::solution Let $x_n=1/(2n\pi+\pi/2)$ and $y_n=1/(2n\pi+3\pi/2)$. Then $x_n\to0$ and $y_n\to0$. Also $f(x_n)=1$ and $f(y_n)=-1$. If a limit existed, both image sequences would converge to the same number. They do not. Hence, $\lim_{x\to0}f(x)$ does not exist. ::: :::exercise 1. Explain why $f(a)$ is not used in the definition of $\lim_{x\to a}f(x)$. 2. Prove that a limit of a mapping, when it exists, is unique. 3. Find $\lim_{x\to0}x\cos(1/x)$. ::: :::answer 1. The limit concerns nearby points distinct from $a$. 2. Use the triangle inequality and take points sufficiently close to $a$. 3. The limit is $0$, since $|x\cos(1/x)|\leq |x|$. ::: :::faq[Frequently Asked Questions] Q: Can a function be discontinuous when the limit exists? A: Yes. If $a\in A$ and the limit exists but is not equal to $f(a)$, the function is not continuous at $a$. Q: Why must $a$ be a limit point? A: Without points of $A$ arbitrarily close to $a$, there is no genuine limiting process. :::