# Prove that a_n tends to 0

#### Alexmahone

##### Active member
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 $\infty$.)

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 $\epsilon<1$

Last edited:

#### ThePerfectHacker

##### Well-known member
Hello,

Let us take $$\varepsilon = 1$$ there is a large enough integer $$N$$ such that if $$n\geq N$$ then, $$\big| \frac{a_n}{b_n} - L \big | < \varepsilon = 1$$. This means that (opening up the absolute value),

$$-1 < \frac{a_n}{b_n} - L < 1 \implies L-1 < \frac{a_n}{b_n} < L+1$$

Now multiply both sides by $$b_n$$, but be careful! If $$b_n$$ happens to be negative the inequality gets reversed. Therefore, this means that, $$(L-1)b_n < a_n < (L+1)b_n$$ if $$b_n > 0$$ and $$n\geq N$$. And that $$(L+1)b_n < a_n < (L-1)b_n$$ if $$b_n < 0$$ and $$n\geq N$$.

Now as $$b_n \to 0$$ by the Squeeze Theorem the sequence $$a_n \to 0$$ also (notice that whether $$b_n$$ is positive or negative does not matter as in both cases these sequences are going to zero anyway).