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

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