posted here https://mathoverflow.net/questions/351749/division-arithmetic-converging
and was told to ask again here with more info.
so given $a_n$→0
and, $\frac{b_n }{a_n}$ → 1
prove $b_n$→0
I tried the following:
$$\lim_{n\rightarrow ∞}\frac{b_n}{a_n}=1$$ $$\lim_{n\rightarrow ∞}\frac{b_n}{a_n}-1=1-1=0$$ $$\lim_{n\rightarrow ∞}(1+\frac{-b_n}{a_n})^{\frac{1}{a_n}^{a_n}}=0$$
so here I got stuck saying $e^{0} = 1 $ and not 0
second way I tried:
given $\varepsilon$ > 0 there is $N_1$ so for every n>$N_1$ there is: $|a_n - a| < \varepsilon$
given $\varepsilon$ > 0 there is $N_2$ so for every n>$N_2$ there is:
$|\frac{b_n}{a_n} - 1| < \frac{\varepsilon}{2}$
$|\frac{b_n-a_n}{a_n}| < \frac{\varepsilon}{2}$
and now I'll make a smaller fraction:
$|\frac{b_n-a_n}{a_n+2}| <|\frac{b_n-a_n}{a_n}| < \frac{\varepsilon}{2}$
I'll define $N_\max$ = max{$N_1, N_2$} so for every n>$N_\max$ there is:
$|\frac{b_n-0}{0+2}| <|\frac{b_n-a_n}{a_n}| < \frac{\varepsilon}{2}$
$|\frac{b_n}{2}| < \frac{\varepsilon}{2}$ $b_n < \frac{2\varepsilon}{2} = \varepsilon$
when n$\rightarrow$ ∞ < $\varepsilon$ so $b_n$ = 0 = a
I just don't know if this is right because I learn calculus alone and there is no answer
Use $$ \lim_{n \to \infty} {b_n}= \lim_{n \to \infty} \left(\frac{b_n}{a_n} \cdot a_n\right) = \left(\lim_{n \to \infty} \frac{b_n}{a_n}\right)\cdot \left({\lim_{n \to \infty} a_n}\right)=1\cdot 0=0 $$