Proof Regarding the Algebraic Limit Theorem

Proof Regarding the Algebraic Limit Theorem
$\begingroup$

Without using the property of the Algebraic Limit Theorem that states, $\lim\limits_{n \to \infty} a_n*b_n = ab$, Prove directly that $\frac{a_n}{b_n} \rightarrow \frac{a}{b}$, if $a_n \to a$ and $b_n \to b$ and b cannot equal zero.

$\endgroup$

1 Answer

$\begingroup$

Since, $b_n \to b$,given $\epsilon>0, $ $\exists N \in \mathbb{N}$, s.t. $\forall n\ge N, |b_n - b|<\epsilon$ and $|a_n-a|\le\epsilon$, then $|b_n|\ge\min \{|b_1|,\cdots,|b_N|,|b|-\epsilon\}=M(say)$ and $|a_n|\le\max \{|a_1|,\cdots,|a_N|,|a|+\epsilon\}=m(say)$.

$|\frac{a_n}{b_n}-\frac{a}{b}|=|\frac{a_n}{b_n}-\frac{a_n}{b}+\frac{a_n}{b}-\frac{a}{b}|\leq|a_n||\frac{1}{b_n}-\frac{1}{b}|+|1/b||a_n-a|\leq \frac{m\epsilon}{bM}+\frac{\epsilon}{b}=\frac{\epsilon}{b}(1+\frac{m}{M})$.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Alexander Ross
Author

Alexander Ross

Alexander Ross has covered the video game industry for a decade, writing deep dives on game design, esports tournaments, VR developments, and gaming culture.