Let \(X\) and \(Y\) be metric spaces and let \(S\subseteq X\) be a subset. If \(f:S\rightarrow Y\) is a function, \(a\in X\) is a cluster point of \(S\text{,}\) and \(y\in Y\text{,}\) we write \(y=\lim_{x\rightarrow a}f(x)\) if for all \(\varepsilon>0\,\exists\,\delta>0\, \forall x\in S\setminus\{{a}\},d_X(a,x)<\delta\implies d_Y(y,f(x))<\varepsilon\text{.}\)
Theorem4.80.Function Limits and Sequence Limits.
Let \(X,Y\) be metric spaces and let \(S\subseteq X\text{.}\) Suppose \(f:S\rightarrow Y\) is a function, \(a\in X\) is a cluster point of \(S\text{,}\) and \(y\in Y\text{.}\) Then \(\lim_{x\rightarrow a}f(x)=y\) iff for all sequences \((x_n)_{n=1}^\infty\subseteq S\setminus\{{a}\}, \lim_{n\rightarrow\infty}x_n=a\implies \lim_{n\rightarrow\infty}f(x_n)=y\text{.}\)
(\(\implies\)) Suppose \(\lim_{x\rightarrow a}f(x)=y\text{.}\) Let \((x_n)_{n=1}^\infty\subseteq S\setminus\{{a}\}, \lim_{n\rightarrow\infty}x_n=a\text{.}\) Fix \(\varepsilon>0\text{.}\) Fix \(\delta>0\) such that if \(x\in S\setminus\{{a}\}\) and \(d(x,a)<\delta\) then \(d(f(x),y)<\varepsilon\text{.}\) Now exists \(N\in\mathbb{Z}_+\) such that for all \(n\in \mathbb{Z}, n\geq N,d(x_n,a)<\delta\text{.}\) Then if \(n\in\mathbb{Z},d(f(x_n),y)<\varepsilon\text{.}\) So \(\lim_{n\rightarrow\infty}f(x_n)=y\text{.}\)
(\(\impliedby\text{,}\) contrapositive) Suppose \(\lim_{x\rightarrow a} f(x)\neq y\text{.}\) Then exists \(\varepsilon>0\) such that for all \(\delta>0\) exists \(x\in S\setminus\{{a}\}, d(x,a)<\delta\) and \(d(f(x),y)\geq \varepsilon\text{.}\) Fix such an \(\varepsilon>0\text{.}\) Then for each \(n\in\mathbb{Z}_+,\exists x_n\in S\setminus\{{a}\}\) such that \(d(x_n,a)<\frac{1}{n},d(f(x_n),y)\geq \varepsilon\text{.}\) Now \((x_n)_{n=1}^\infty\subseteq S\setminus\{{a}\},\lim_{n\rightarrow\infty}x_n=a\text{,}\) and \((f(x_n))_{n=1}^\infty\) does not converge to \(y\text{.}\) (remark “this is maybe a bit pedantic” \(\lim_{n\rightarrow\infty}f(x_n)\neq y\) doesn't really make sense since the left hand side doesn't have to exist. If you're writing \(\lim_{n\rightarrow\infty}f(x_n)\) you're assuming it exists.)
Example4.81.
\(f:\mathbb{R}^2\setminus\{{(0,0)}\}\rightarrow\mathbb{R}\) with \(f(x,y)=\frac{(x+y)^2}{x^2+y^2}\text{.}\) We'll show \(\lim_{(x,y)\rightarrow (0,0)}f(x,y)\) does not exist by constructing sequences \(((x_n,y_n))_{n=1}^\infty\) and \(((x_n',y_n'))_{n=1}^\infty\) converging in \(\mathbb{R}^2\setminus\{{0}\}\) so that the sequences \((f(x_n,y_n))_{n=1}^\infty\) and \((f(x_n',y_n'))_{n=1}^\infty\) converge to different values.
Take \((x_n,y_n)=(\frac{1}{n},0)\) then \(\lim_{n\rightarrow\infty}f(x_n,y_n)=\lim_{n\rightarrow\infty}\frac{(\frac{1}{n}+0)^2}{\frac{1}{n}^2+0^2}=1\text{,}\) and take \((x_n',y_n')=(\frac{1}{n},\frac{1}{n})\) then \(\lim_{n\rightarrow\infty}f(x_n',y_n')=\lim_{n\rightarrow\infty}\frac{(\frac{1}{n}+\frac{1}{n})^2}{\frac{1}{n}^2+\frac{1}{n}^2}=2\text{.}\) So \(\lim_{(x,y)\rightarrow(0,0)}f(x,y)\) does not exist.
(If \(\lim_{(x,y)\rightarrow(0,0)}f(x,y)=L,\) then for any sequence \(((x_n,y_n))_{n=1}^\infty\) converging to \((0,0)\text{,}\)
Define \(f:(0,\infty)\rightarrow \mathbb{R}\) by \(f(x)=\sin(\frac{1}{x}\) (oscillates wildly near origin). Called “topologist's sine curve”. For any \(y\in[-1,1]\) there is a sequence \((x_n)_{n=1}^\infty\) such that \(\lim_{n\rightarrow\infty}x_n=0\) and \(\lim_{n\rightarrow\infty} f(x_n)=y\text{.}\) In particular \(\lim_{x\rightarrow 0}f(x)\) does not exist.
