Using Basics Limit Arithmetics

Given $\lim \limits_{n\to \infty}a_n=L$ and $\lim \limits_{n\to \infty}\frac{a_n}{b_n}=1$ then $\lim \limits_{n\to \infty}b_n=L$

I think that this claim is true:

$$\lim \limits_{n\to \infty}a_n=L$$

$$\lim \limits_{n\to \infty}\frac{a_n}{b_n}=1$$

using limits arithmetics I can say:

$$\lim \limits_{n\to \infty}\frac{a_n}{b_n}=\frac{\lim \limits_{n\to \infty}a_n}{\lim \limits_{n\to \infty}b_n}=1$$

$$\frac{\lim \limits_{n\to \infty}a_n}{\lim \limits_{n\to \infty}b_n}=1$$

$$\lim \limits_{n\to \infty}b_n\cdot 1=\lim \limits_{n\to \infty}a_n$$

$$\lim \limits_{n\to \infty}b_n=L.$$

is my proof correct?

-

I'm not sure it's correct. In the first line after "I can say", it seems that you are assuming $\lim\limits_{n\rightarrow\infty}b_n=L$, which is what you are trying to prove.

However, you can use the following fact: Suppose $\lim\limits_{n\rightarrow\infty}a_n= a$ and that $\lim\limits_{n\rightarrow\infty}c_n= c$. Then if $c\ne0$, the limit of the quotient ${a_n\over c_n}$ as $n$ tends to infinity exists and $\lim\limits_{n\rightarrow\infty}{a_n\over c_n}={a\over c }$.

Apply this with $c_n={a_n\over b_n}$.

-
hmm, I show just in the last line that $\lim\limits_{n\rightarrow\infty}b_n=L$, I just use multiplication throught the proof to show it. –  Anonymous Apr 1 '12 at 16:40
@Anonymous but you can't say that $\lim \limits_{n\to \infty}\frac{a_n}{b_n}=\frac{\lim \limits_{n\to \infty}a_n}{\lim \limits_{n\to \infty}b_n}$ unless you know that both $\lim\limits_{n\rightarrow\infty} a_n$ and $\lim\limits_{n\rightarrow\infty} b_n$ exist with $\lim\limits_{n\rightarrow\infty} b_n\ne0$. At the start, you do not even know that $\lim\limits_{n\rightarrow\infty} b_n$ exists. –  David Mitra Apr 1 '12 at 16:46
Great so here is the proof: $\lim \limits_{n\to \infty}\frac{a_n}{c_n}=\lim \limits_{n\to \infty}\frac{a_n * b_n}{a_n}=\lim \limits_{n\to \infty}b_n$ and on the other hand we know $\lim \limits_{n\to \infty}\frac{a_n}{c_n}=\frac{\lim \limits_{n\to \infty}a_n}{\lim \limits_{n\to \infty}c_n}=L$ and therefore $\lim \limits_{n\to \infty}b_n=L$. am I correct? –  Anonymous Apr 1 '12 at 17:08
@Anonymous Yes, that looks good. But I would write (I think it would be a bit more clear) the "other hand" first and then state "but ${a_n\over c_n}=b_n$ when $b_n\ne0$, so $\lim\limits_{n\rightarrow\infty} b_n=\lim\limits_{n\rightarrow\infty}{a_n\over c_n}=L$". –  David Mitra Apr 1 '12 at 17:12
Awesome sauce, thanks a lot :) –  Anonymous Apr 1 '12 at 17:17

Given $\lim \limits_{n\to \infty}a_n=L$ and $\lim \limits_{n\to \infty}\frac{a_n}{b_n}=1$ then $\lim \limits_{n\to \infty}b_n=L$

Try $\lim \limits_{n\to \infty}\frac{a_n}{b_n}\frac1{a_n}=\frac1L$

-
what's the difference between the two proofs? –  Anonymous Apr 1 '12 at 16:10
They're not different, I posted mine before the other guy. It's all the same. –  Steven-Owen Apr 2 '12 at 0:12