---- > [!definition] Definition. ([[function limit]]) > Let $f:X \to Y$ be a function between [[topological space|topological spaces]], and let $x_{0} \in X$ be a [[limit point]] of $X$. We say $\lim _{x \to x_{0}}f(x)=L$ if for every [[neighborhood]] $V \ni L$ there exists a [[neighborhood]] $U$ of $x_{0}$ such that $f(U-\{ x_{0} \}) \subset V$. > If $Y$ is [[Hausdorff space|Hausdorff]], then $L$ is unique. ^definition > [!basicnonexample] Basic Nonexistence Criterion. > If $\lim_{x \to x_{0}}f(x)=L$, then in particular for any curve $\gamma:I \to X$ from $x$ to $x_{0}$[^1] it holds $\lim_{x \to x_{0}}(f \circ \gamma)(x)=L$. Thus if $Y$ is Hausdorff so that limits are unique, to show nonexistence of $L$ it suffices to witness two curves $\gamma_{1},\gamma_{2}$ such that $\lim_{x \to x_{0}}(f \circ \gamma_{1})(x) \neq \lim_{x \to x_{0}}(f \circ \gamma_{2})(x)$. ^nonexample [^1]: We should assume also that $\gamma(t) \neq x_{0}$ for $t \neq t_{0}$ sufficiently close to $t_{0}$. > [!note] Remark. > The definition would become vacuous if it were not enforced that $x_{0}$ is a [[limit point]] of $X$. Indeed, if $x_{0}$ were an isolated point, then there would exist a neighborhood $U \ni x_{0}$ for which $X \cap (U-\{ x_{0} \})=\emptyset$, meaning $U-\{ x_{0} \}$ is empty and so $f(U-\{ x_{0} \}) =f(\emptyset)=\emptyset \subset V$ for any open $V \subset Y$: every $L \in Y$ would be a valid limit! ^note > [!basicproperties] > - If $Y$ is [[Hausdorff space|Hausdorff]], then $L$ is unique. > - For [[topological space|topologys]] [[topology generated by a basis|generated by]] a [[basis for a topology|basis]], it is enough to check assuming $V,U$ are basic open sets. ^properties Things to find and link to: - [ ] Hausdorff -> unique - [ ] sequence characterization proof - [ ] general characterization with nets > [!equivalence] Sequential characterization for [[first-countable space|first-countable spaces]]. > If $X,Y$ are [[first-countable space|first-countable]] then $\lim_{x \to x_{0}}f(x)=L$ if and only if for every [[sequence]] $(x_{n}) \to x_{0}$ in $X$ one has $f(x_{n}) \to L$ in $Y$. ^equivalence > [!equivalence] $\varepsilon \text{-}\delta$ characterization for metric spaces. > If $X,Y$ are [[metric space|metric spaces]], then $\lim_{x \to x_{0}}f(x)=L$ if and only if for all $\varepsilon>0$, there exists $\delta>0$ such that $0< d(x, x_{0})<\varepsilon \implies d\big( f(x), L \big)<\delta.$ ^equivalence ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` a#analysis/real-analysis # definition Let $Z, Y$ be [[metric space]]s . Suppose $f: x \subset Z \to Y$. Suppose $p$ is a [[limit point]] of $X$ and $q \in Y$. We say $f(x) \to q$ as $x \to p$ , or $\lim_{ x \to p }f(x)=q$ if $\forall \varepsilon >0$ $\exists \delta>0$ s.t. $0<d_{X}(x,p) < \delta \implies d_{Y}(f(x), q) < \varepsilon$ that is, $x \in U_{X}(p;\delta) \backslash {\{p\}} \implies f(x) \in U_{Y}(q, \varepsilon)$ "*as long as x is close to p, f(x) should be close to q*" #### Remark If $f(x)$ converges to $q$ as $x \to p$ , then the limit is unique - Proved in [[Math 297]] using the [[triangle inequality]] The following intuitive equivalency holds: $\lim_{ x \to p }f(x)=q \iff \lim_{ d_{x}(x,p) \to 0 }d_{Y}(f(x),q)=0$ # Theorems For $\vec f: X \subset Z \to \mathbb{R}^n$ with $\vec f(x) = \begin{bmatrix} f_{1}(x) \\ f_{2}(x) \\ \vdots \\ f_{n}(x) \\ \end{bmatrix}$, we have component-wise convergence: $\lim_{ x \to q } \vec f(x) = \vec q \iff \lim_{ x \to p }f_{i}(x) = q_{i} \text{ for } 1 \leq i \leq n $ Some more limit properties: if $f,g: X \subset Z \to \mathbb{R}$ and $f(x)\to a$ , $g(x)\to b$ as $x \to p$, then 1. $\alpha f(x) + \beta g(x) \to \alpha a + \beta b$ 2. $f(x)g(x) \to ab$ 3. $\frac{f(x)}{g(x)} \to \frac{a}{b}$ if $b\neq a$ #notFormatted