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If the limit of one sequence $\{a_n\}$ is zero and the limit of another sequence $\{b_n\}$ is also zero does that mean that $\displaystyle\lim_{n\to\infty}(a_n/b_n) = 1$?

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What have you tried? What happens when $a_n=b_n^2$? What happens when $b_n=2a_n$? –  Thomas Andrews Jun 26 at 15:06
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Sequence $\frac{1}{n} $ and $\frac{1}{2n}$ –  Swapnil Tripathi Jun 26 at 15:09
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We can conclude essentially nothing about the behaviour of $\frac{a_n}{b_n}$. –  André Nicolas Jun 26 at 15:13
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So for shorthand we say: $0/0$ is an indeterminate form. –  GEdgar Jun 26 at 15:26

5 Answers 5

Given $b_n$ whose limit is 0, we can choose $a_n$ to make the limit:

  • 1: set $a_n = b_n$

  • 0: set $a_n = b_n^2$, or even $a_n = 0$

  • Some other constant c: set $a_n = cb_n$

  • Undefined: $a_n = (-1)^nb_n$

  • Infinite: $a_n = \sqrt{|b_n|}$

So we can make it basically anything.

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If we want an example such that the limit exists: Let $r \in \Bbb{R}$ be your favorite number, suppose that $b_n \to 0$, and let $a_n = r b_n$. Then $a_n \to 0$ and $a_n/b_n \to r$.

If we want an example such that the limit does not exist: Let $b_n \to 0$ and $\{c_n\}$ be your favorite bounded sequence such that $\lim c_n$ does not exist. Let $a_n = c_n b_n$, then $a_n \to 0$ and $a_n/b_n = c_n$, and the limit does not exist.

But, don't consider this a disappointing result! We would be in a much different world if this was not the case; for example, imagine how boring L'Hopital's rule would be.

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Of course not.

  • $a_n=\dfrac{(-1)^n}{n}$,
  • $b_n=\dfrac{1}{n}$,
  • Hence, $r_n=\dfrac{a_n}{b_n}=(-1)^n$ and the limit of $r_n$ does not even exist.
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Let $a_n=1/n$, $b_n=1/n^2$. What can you say on the ratio?

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No. Let $a_n = \dfrac 1 n$ and $b_n = \dfrac 1 {n^2}$.

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