Here we present the proof of assertions 3. and 4. of the following proposition. The proofs of 1. and 2. will be presented in class and here they are assumed. The proofs presented here will not be presented in class.

Proposition 3.1.4 (Calculus of Limits)
Suppose that $f$ and $g$ are two functions $\mathbb{R}\rightarrow \mathbb{R}$, and that for some $a\in\mathbb{R}$ we have $\lim_{x\rightarrow a}f(x)=p$ , and $\lim_{x\rightarrow a}g(x)=q$.

for some $p,q\in\mathbb{R}$. Then

1. $\lim_{x\rightarrow a} (f(x)+g(x))=p+q$.
2. If $k\in\mathbb{R}$, $\lim_{x\rightarrow a} kf(x)=kp$.
3. $\lim_{x\rightarrow a} (f(x)g(x))=pq$.
4. If $q\neq0$, $\lim_{x\rightarrow a} (f(x)/g(x))=p/q$.
5. If $n\in\mathbb{N}$, and $p>0$ then $\lim_{x\rightarrow a} \sqrt[n]{f(x)}=\sqrt[n]{p}$.

Proof of 3: For a constant function, $\lim_{x\rightarrow} k=k$ [Example 3.1.3 (2)]. By 1. $\lim_{x\rightarrow a}(f(x)-p)=p-p=0$, and $\lim_{x\rightarrow a}(g(x)-q) =q-q=0.$

Now, let $\varepsilon>0$.  Then there is a $\delta_1>0$ and a $\delta_2>0$ such that, $|(f(x)-p)-0|< \sqrt{\varepsilon}$ whenever $0<|x-a|<\delta_1$, and $|(g(x)-q)-0|< \sqrt{\varepsilon}$ whenever $0<|x-a|<\delta_2$

Choose $\delta:=\min\{\delta_1,\delta_2\}$.  If $0<|x-a|<\delta$ we then get, $\left|(f(x)-p)(g(x)-q)-0\right|=|f(x)-p||g(x)-q|<\sqrt{\varepsilon}\sqrt{\varepsilon}<\varepsilon$

Now $(f(x)-p)(g(x)-q)=f(x)g(x)-qf(x)-pg(x)+pq$ $\Rightarrow$ $f(x)g(x)= (f(x)-p) (g(x)-q)+qf(x)+pg(x)-pq$

By the Sum Rule (1.): $\lim_{x\rightarrow a} f(x)g(x)=\underbrace{\lim_{x\rightarrow a}(f(x)-p)(g(x)-q)}_{=0}$ $+\lim_{x\rightarrow a}qf(x)+\lim_{x\rightarrow a} pg(x)-pq$ $=0+qp+pq-pq=pq$.

The last line guaranteed by 2.

Proof of 4: Let $\varepsilon>0$, and assume that $p\neq 0$. Set $\varepsilon_f:=\frac{\varepsilon|q|}{4}>0$ and $\varepsilon_g:=\min\left\{\frac{\varepsilon|q|^2}{4|p|},\frac{|q|}{2}\right\}>0.$

By hypothesis there are $\delta_f>0$ and $\delta_g>0$ such that $0<|x-a|<\delta_{f}$ $\Rightarrow$ $|f(x)-p| < \varepsilon_{f}$ and $0<|x-a|<\delta_g$ $\Rightarrow$ $|g(x)-q|<\varepsilon_g$ $(\star)$

By the reverse triangle inequality we have $|g(x)|=|q-(q-g(x))|\geq |q|-|q-g(x)|$

and so if $0<|x-a|<\delta_g$ then $|g(x)-q|<\varepsilon_g\leq |q|/2$, and hence $|g(x)|\geq |q|-\frac{|q|}{2}=\frac{|q|}{2}>0$ ( $\dag$)

Define $\delta:=\min\{\delta_f,\delta_g\}$, so in particular $\delta>0$. Now $\frac{f(x)}{g(x)}-\frac{p}{q}=\frac{qf(x)\overbrace{-qp+qp}^{=0}-pg(x)}{qg(x)}$

Thus if $0<|x-a|<\delta$, then the inequalities ( $\dag$) and ( $\star$) hold so $|q(f(x)-p)-p(g(x)-q)|\leq |q||f(x)-p|+|p||g(x)-q|$ $\leq |q|\varepsilon_f+|p|\varepsilon_g$ $\leq |q|\frac{\varepsilon |q|}{4}+|p|\frac{\varepsilon |q|^2}{4|p|}=\frac{\varepsilon |q|^2}{2}$

and $|q g(x)|=|q| |g(x)|\geq |q|^2/2$

So finally, we have $0<|x-a|<\delta\Rightarrow \left|\frac{f(x)}{g(x)}-\frac{p}{q}\right|<\frac{\varepsilon|q|^2/}{|q|^2/2}=\varepsilon$

as required.

The case where $p=0$  is done by setting $\varepsilon_f=\varepsilon|q|/2$ and $\varepsilon_g=|q|/2$ and following the same steps (and is left as an exercise) $\bullet$