# Equivalence of limits? [closed]

Let $(a_n)$ and $(b_n)$ be two sequences. Is it always right to say that $$\lim_{n\to\infty}(a_n-b_n)=0 \iff \lim_{n\to\infty}\frac{a_n}{b_n}=1$$ ?

• Provided $\lim_{n\to\infty}(a_n) \neq 0$ – Donald Splutterwit Aug 4 '17 at 14:13
• It would be more convincing to think that $$\lim_{n \to \infty} |a_n - b_n| = 0 \iff \lim_{n \to \infty} \frac{|a_n|}{|b_n|} = 1$$ But this still isn't true. – Dionel Jaime Aug 4 '17 at 14:22
• – Hans Lundmark Aug 4 '17 at 17:54

## 5 Answers

As others have shown this isn't true in general. In their examples, the sequences $a_n$ and $b_n$ either both tend to $0$ or to $\infty$. What happens in other cases?

Suppose that $\lim b_n = L$ is finite and $\lim \frac{a_n}{b_n} = 1$. Then, $$\left| a_n - b_n \right| = |b_n| \left| \frac{a_n - b_n}{b_n} \right| = |b_n| \left| \frac{a_n}{b_n} - 1 \right| \to |L| \cdot 0 = 0$$ Thus, $\lim (a_n-b_n) = 0.$

In the same way, if $\lim b_n = L \neq 0$ and $\lim (a_n-b_n) = 0$ then $$\left| \frac{a_n}{b_n} - 1 \right| = \left| \frac{a_n-b_n}{b_n} \right| = \frac{1}{|b_n|} \left| a_n-b_n \right| \to \frac{1}{|L|} \cdot 0 = 0$$ Thus, $\lim \frac{a_n}{b_n} = 1.$

Therefore, if $\lim b_n$ exists and is neither zero or infinite, then we do have equivalence. In this case, of course $\lim a_n = \lim b_n = L.$

No. Let $a_n=n$, and $b_n=n+1.$

• This quotient still converges to 1 ... – Dionel Jaime Aug 4 '17 at 14:15
• @DionelJaime um what? We have $a_n-b_n$ converges to $-1$ and $\frac{a_n}{b_n}$ converges to $1$. – edm Aug 4 '17 at 14:17
• Ahh I didn't notice his double implication, sorry ... – Dionel Jaime Aug 4 '17 at 14:17

$$a_n=\frac 1{n^2}, b_n =\frac{1}{n}$$

Then $$\lim (a_n-b_n)=0$$ But $$\lim \frac{a_n}{b_n}=\lim \frac{1}{n}=0$$

If only the world was that simple.

Let $a_n = \frac{1}{2^{n-1}}$ and $b_n = \frac{1}{2^{n}}$

If $a_n=\frac{\sin{n}}{n}, b_n=\frac{1}{n}$, then: $$\lim_\limits{n\to\infty} (a_n-b_n)=0,$$ $$\lim_\limits{n\to\infty}\frac{a_n}{b_n}=\lim_\limits{n\to\infty}\sin{n},$$ which does not exist.