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#### Alexmahone

##### Active member

- Jan 26, 2012

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Prove: if $\frac{a_n}{b_n}\to L$, $b_n\neq 0$ for any $n$, and $b_n\to 0$, then $a_n\to 0$. ($L$ represents a finite number, not $\infty$.)

Given $\epsilon>0$,

$\left|\frac{a_n}{b_n}-L\right|<\epsilon$ and $|b_n|<\epsilon$

How do I proceed?

---------- Post added at 11:43 AM ---------- Previous post was at 10:55 AM ----------

Let me give it a shot.

$\left|\frac{a_n}{b_n}\right|=\left|\frac{a_n}{b_n}-L+L\right|\le\left|\frac{a_n}{b_n}-L\right|+|L|<\epsilon+|L|$

$|a_n|<|b_n|(\epsilon+|L|)<\epsilon(\epsilon+|L|)<\epsilon(1+|L|)$

since we may assume $\epsilon<1$

**My working**:Given $\epsilon>0$,

$\left|\frac{a_n}{b_n}-L\right|<\epsilon$ and $|b_n|<\epsilon$

How do I proceed?

---------- Post added at 11:43 AM ---------- Previous post was at 10:55 AM ----------

Let me give it a shot.

$\left|\frac{a_n}{b_n}\right|=\left|\frac{a_n}{b_n}-L+L\right|\le\left|\frac{a_n}{b_n}-L\right|+|L|<\epsilon+|L|$

$|a_n|<|b_n|(\epsilon+|L|)<\epsilon(\epsilon+|L|)<\epsilon(1+|L|)$

since we may assume $\epsilon<1$

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