Real analysis exercise asked in many math groups (Introduction to Sequences using Epsilon -limit definition)

Exercise:

Consider two sequences ${{\left( {{a}_{n}} \right)}_{n}}\,\,\And \,\,\,\,\,{{\left( {{b}_{n}} \right)}_{n}}$ converges to $a$ and $b$ respectively

Show that, ${{a}_{n}}{{b}_{n}}\xrightarrow[n\to \infty ]{}ab$ and $\frac{{{a}_{n}}}{{{b}_{n}}}\xrightarrow[n\to \infty ]{}\frac{a}{b}$ , $b\ne 0$

Solution: we have ${{a}_{n}}{{b}_{n}}\to ab\Leftrightarrow \left| {{a}_{n}}{{b}_{n}}-ab \right|\to 0$ ??

We have $\left| {{a}_{n}}{{b}_{n}}-ab \right|=\left| {{a}_{n}}{{b}_{n}}+a{{b}_{n}}-a{{b}_{n}}-ab \right|$

$\left| {{b}_{n}}\left( {{a}_{n}}-a \right)+a\left( {{b}_{n}}-b \right) \right|\le \left| {{b}_{n}} \right|\left| {{a}_{n}}-a \right|+\left| a \right|\left| {{b}_{n}}-b \right|$

Since ${{b}_{n}}$ converges so $\exists \,A>0$ such that $\left| {{b}_{n}} \right|<A$ $\forall n$

Take $\varepsilon >0$ , there is ${{N}_{1}}$ such that if $n\ge {{N}_{1}}$ then $\left| {{a}_{n}}-a \right|<\frac{\varepsilon }{2A}$

Also there is ${{N}_{2}}$ such that if$n\ge {{N}_{2}}$ then $\left| {{b}_{n}}-b \right|<\frac{\varepsilon }{2a}$

So $\left| {{a}_{n}}{{b}_{n}}-ab \right|\le A\frac{\varepsilon }{2A}+a\frac{\varepsilon }{2a}=\frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon $

It’s enough to prove that $\frac{1}{{{b}_{n}}}$ converges to $\frac{1}{b}$ using previous result

      $\left| \frac{1}{{{b}_{n}}}-\frac{1}{b} \right|=\left| \frac{b-{{b}_{n}}}{{{b}_{n}}b} \right|=\left| \frac{{{b}_{n}}-b}{{{b}_{n}}b} \right|=\frac{\left| {{b}_{n}}-b \right|}{\left| b \right|\left| {{b}_{n}} \right|}$

Since ${{b}_{n}}$ converges to $b$ there is ${{N}_{1}}$ such that if $n\ge {{N}_{1}}$ then $\left| {{b}_{n}}-b \right|<\frac{\left| b \right|}{2}$

Let $\varepsilon =\frac{\left| b \right|}{2}>0$ as $b\ne 0$

Observe that $\left| b \right|=\left| b+{{b}_{n}}-{{b}_{n}} \right|=\left| \left( {{b}_{n}}-b \right)+{{b}_{n}} \right|\le \left| {{b}_{n}}-b \right|+\left| {{b}_{n}} \right|=\frac{\left| b \right|}{2}+\left| {{b}_{n}} \right|$

$\Leftrightarrow \frac{\left| b \right|}{2}<\left| {{b}_{n}} \right|$ for all $n\ge {{N}_{1}}$ $\Leftrightarrow \frac{{{b}^{2}}}{2}<\left| b \right|\left| {{b}_{n}} \right|\Leftrightarrow \frac{1}{\left| b \right|\left| {{b}_{n}} \right|}<\frac{2}{{{b}^{2}}}$

Also on the other hand we have ${{b}_{n}}$ converges to $b$

So for any $\varepsilon >0$ there is ${{N}_{2}}$ such that if $n\ge {{N}_{2}}$ then $\left| {{b}_{n}}-b \right|\le \frac{\varepsilon {{b}^{2}}}{2}$

Hence for $n\ge \max \left( {{N}_{1}},{{N}_{2}} \right)$ we have $\left| \frac{1}{{{b}_{n}}}-\frac{1}{b} \right|=\frac{\left| {{b}_{n}}-b \right|}{\left| b \right|\left| {{b}_{n}} \right|}=\frac{1}{\left| b \right|\left| {{b}_{n}} \right|}\left| {{b}_{n}}-b \right|<\frac{2}{{{b}^{2}}}\times \frac{\varepsilon {{b}^{2}}}{2}=\varepsilon $

Using previous result the proof is complete