Limit theorem. Let $\lim _{n \rightarrow \infty} a_{n}=\alpha, \lim _{n \rightarrow \infty} b_{n}=\beta$. $\alpha, \beta \in \mathbb{R}$.
Then the followings hold:
(a) $\lim _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\alpha+\beta$,
(b) $\lim _{n \rightarrow \infty}\left(k a_{n}\right)=k \alpha$ for any $k \in \mathbb{R}$,
(c) $\lim _{n \rightarrow \infty}\left(a_{n}-b_{n}\right)=\alpha-\beta$,
(d) $\lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right)=\alpha \beta$,
(e) $\lim _{n \rightarrow \infty}\left(\frac{a_{n}}{b_{n}}\right)=\frac{\alpha}{\beta}$ provided $b_{n}, \beta \neq 0$ for any $n \in \mathbb{N}$.
Proof. (a) Let $c_{n}=a_{n}+b_{n}$.
Need to Show (NTS): $\lim _{n \rightarrow \infty} c_n=\alpha+\beta$.
Let $\varepsilon>0$ be fixed.
Since $\lim _{n \rightarrow \infty} a_{n}=\alpha_{1} \quad \lim _{n \rightarrow \infty} b_{n}=\beta$,
there exists $N_{\varepsilon}, N_{\varepsilon}^{\prime} \in \mathbb{N}$ such that
for any $n \geq N_{\varepsilon}, \quad\left|a_{n}-\alpha\right|<\varepsilon / 2$,
for any $n \geq N_{\varepsilon}^{\prime}, \quad\left|b_{n}-\beta\right|<\varepsilon / 2$.
Take $N=\max \left\{N_{\varepsilon}, N_{\varepsilon}^{\prime}\right\}$.
Then $n \geq N$,
$$
\begin{aligned}
& \mid\left(c_n-(\alpha+\beta)|=| a_{n}+b_{n}-(\alpha+\beta) \mid\right. \\
& =\left|a_{n}-\alpha+b_{n}-\beta\right| \stackrel{\text { Tri.Ineq }}{\leq}\left|a_{n}-\alpha\right|+\left|b_{n}-\beta\right| <\varepsilon / 2+\varepsilon / 2 \\
& =\varepsilon
\end{aligned}
$$
Hence for each $\varepsilon>0$, there exists $N \in \mathbb{N}$ s.t for any $n \geq N,|c_n-(\alpha+\beta)|<\varepsilon$
$$
\Leftrightarrow \lim _{n \rightarrow \infty} c_{n}=\lim _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\alpha+\beta.
$$
(b), (c): Exercise.
(d) Let $c_{n}=a_{n} b_{n}$.
$$
\begin{aligned}
& \text { NTS: } \lim _{n \rightarrow \infty} c_{n}=\alpha \beta \\
& \text { Idea } \left.\left|a_{n} b_{n}-\alpha \beta\right|=\mid a_{n}-\alpha\right) b_{n}+\alpha b_{n}-\alpha \beta \mid \\
& =\left|\left(a_{n}-\alpha\right) b_{n}+\alpha\left(b_{n}-\beta\right)\right| \\
& \leq\left|a_{n}-\alpha\right| \cdot\left|b_{n}\right|+|\alpha| b_{n}-\beta \\
& \leq M\left|a_{n}-\alpha\right| +|\alpha| b_{n}-\beta<\varepsilon.
\end{aligned}
$$
Let $\varepsilon>0$ be fixed.
Since $\lim _{n \rightarrow \infty} b_{n}=\beta$, there exists $M>0$ such that
$\left|b_{n}\right| \leq M$ for any $n \in \mathbb{N}$ (this follows from previous lecture).
Since $\lim _{n \rightarrow \infty} a_{n}=\alpha, \lim _{n \rightarrow \infty} b_{n}=\beta$, there exists $N_{\varepsilon}, N_{\varepsilon}^{\prime} \in \mathbb{N}$ such that
for any $n \geq N \varepsilon,\left|a_{n}-x\right|<\frac{\varepsilon}{2 M}$, for any $n \geq N_{\varepsilon}^{\prime},\left|b_{n}-\beta\right|<\frac{\varepsilon}{2|\alpha|}$.
Put $N=max(N_{\varepsilon}^{\prime},N_{\varepsilon})$.
Then for any $n \geq N$,
$$
\begin{aligned}
\left|c_{n}-\alpha \beta\right|=\left|a_{n} b_{n}-\alpha \beta\right| & \leq M \cdot\left|a_{n}-\alpha\right| +|\alpha| \cdot\left|b_{n}-\beta\right| \\
& <M \cdot \frac{\varepsilon}{2 M}+|\alpha| \cdot \frac{\varepsilon}{2|\alpha|} \\
& =\varepsilon.
\end{aligned}
$$
Hence $\forall \varepsilon>0, \exists N \in \mathbb{N}$ st $\forall n \geq N,\left|c_{n}-\alpha \beta\right|<\varepsilon$.
(e) Sketch of idea
Need: $\left|\frac{a_{n}}{b_{n}}-\frac{\alpha}{\beta}\right|<\varepsilon . \quad\left|a_{n}-\alpha\right|,\left|b_{n}-\beta\right|<\varepsilon$
$$
\left|\frac{\beta a_{n}-\alpha b_{n}}{\beta b_{n}}\right|=\frac{1}{\left|\beta b_{n}\right|} \mid \beta\left(a_{n}-\alpha)-\alpha\left(b_{n}-\beta\right) \mid=(*)\right.
$$
From Triangle. Inequality, $|x+y| \leq|x|+|y|$
$$
\begin{aligned}
& \Leftrightarrow|y|=|y-x+x| \leq|y-x|+|x| \\
& \Leftrightarrow|y|-|x| \leq|y-x| \\
& y=\beta, x=b_{n} . \\
& \Rightarrow|\beta|-\left|b_{n}\right| \leq\left|b_{n}-\beta\right|<\varepsilon=\frac{1}{2}|\beta| \\
& \Rightarrow|\beta|-\left|b_{n}\right|<\frac{1}{2}|\beta| \Leftrightarrow\left|b_{n}\right|>\frac{1}{2}|\beta| \\
& \Leftrightarrow \frac{1}{\left|b_{n}\right|}<\frac{2}{|\beta|}.
\end{aligned}
$$
$$
\begin{aligned}
(*) & \leq \frac{2}{|\beta|^{2}} \times\left|\beta\left(a_{n}-\alpha\right)-\alpha\left(b_{n}-\beta\right)\right| \\
& =\frac{2}{\beta^{2}} \times\left|\beta\left(a_{n}-\alpha\right)+\alpha\left(\beta-b_{n}\right)\right| \\
& \leq \frac{2}{\mid \beta^{2}}\left(\left|\beta \cdot\left(a_{n}-\alpha\right)\right|+\left|\alpha \cdot\left(\beta-b_{n}\right)\right|\right)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
\end{aligned}
$$
Now, let us prove (e).
Fix $\varepsilon>0$.
NTS: $\exists N \in \mathbb{N}$ s.t $\forall_{n} \geq N,\left|\frac{a_{n}}{b_{n}}-\frac{\alpha}{\beta}\right|<\varepsilon$
Take $\varepsilon_{1}=\frac{1}{2}|\beta|>0$.
Since $\lim _{n \rightarrow \infty} b_{n}=l, \exists N_{1} \in \mathbb{N}$ set $\quad \forall n \geq N_{1},\left|k_{n}-\beta\right|<\varepsilon_{1}$.
$$
\begin{aligned}
& \Rightarrow \forall n \geq N_{1},|\beta|-\left|b_{n}\right| \leq\left|b_{n}-\beta\right|<\varepsilon_{1}=\frac{1}{2}|\beta| \\
& \Leftrightarrow \forall n \geq N_{1},\left|b_{n}\right|>\frac{1}{2}|\beta| \\
& \Leftrightarrow \forall n \geq N_{1}, \frac{1}{\left|b_{n}\right|}<\frac{2}{|\beta|}
\end{aligned}
$$
Next, take $\varepsilon_{2}=\frac{\varepsilon}{2} \cdot \frac{1}{|\beta|} \cdot \frac{\mid \beta^{2}}{2}$,
$\varepsilon_{3}=\frac{\varepsilon}{2} \frac{1}{(\alpha \mid} \cdot \frac{|\beta|^{2}}{2}$.
From the assumption $\lim_{n \rightarrow \infty} a_{n}=x, \lim _{n \rightarrow \infty} b_{n}=\beta$, $\exists N_{2}, N_{3} \in \mathbb{N}$ such that
$$
\begin{aligned}
& \forall n \geq N_{2},\left|a_{n}-\alpha\right|<\varepsilon_{2} \\
& \forall n \geq N_{3},\left|b_{n}-\beta\right|<\varepsilon_{3}
\end{aligned}
$$
Choose $N= max \left\{N_{1}, N_{2}, N_{3}\right\}$.
Then $ n \geq N$,
$$
\begin{aligned}
& \left|\frac{a_{n}}{b_{n}}-\frac{\alpha}{\beta}\right|=\left|\frac{\beta a_{n}-\alpha b_{n}}{\beta b_{n}}\right|=\left|\frac{\beta\left(a_{n}-\alpha\right)+\alpha\left(\beta-b_{n}\right)}{\beta t_{n}}\right| \\
& N_{1} \frac{z}{|\beta|^{2}}\left|\beta\left(a_{n}-\alpha\right)+\alpha\left(\beta-t_{n}\right)\right| \\
& \leq \frac{2}{|\ell|^{2}}\left(\left|\beta\left(a_{n}-\alpha\right)\right|+\left|\alpha \cdot\left(\beta-b_{n}\right)\right|\right) \\
& <\frac{2}{|\beta|^{2}} \times\left(|\beta| \cdot \varepsilon_{2}+|\alpha|-\varepsilon_{3}\right) \\
& =\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .
\end{aligned}
$$
Thus, $\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\frac{\alpha}{\beta}$.
Exercise. Suppose $ \lim _{n \rightarrow \infty} a_{n}=\alpha \text { and } \lim _{n \rightarrow \infty} a_{n}=\beta $
Show that $\alpha=\beta$.