Let An Bn and Cn be sequences satisfying An<=Bn<=Cn

[steven187]

hello all

been workin on this problem:
let An Bn and Cn be sequences satisfying
An<=Bn<=Cn for all n an element of the natural numbers
suppose that An->x and Cn->x, where x is a real number show that Bn->x
this is how i did it


A_n\le B_n\le C_n \forall n\epsilon N

A_n\longrightarrow x,C_n\longrightarrow x\ \forall x\epsilon \Re

\lim_{n\to\infty}A_n\le\lim_{n\to\infty}B_n\le\lim_{n\to\infty}C_n

x\le\lim_{n\to\infty}B_n\le x

therefore by the squeeze theorem B_n\longrightarrow x

would this be correct, and are there any other ways of proving it?

thanxs

 

[quasar987]

What you wrote it confusing, for many reasons. First, the question

"Let An Bn and Cn be sequences satisfying An<=Bn<=Cn for all n an element of the natural numbers. Suppose that An->x and Cn->x, where x is a real number. Show that Bn->x."

must mean "prove the squeeze theorem". Otherwise, Bn->x is just the statement of the squeeze theorem and there's nothing to show at all.

Secondly, line #2 makes no sense (because if the limit of An exists, it is unique), but it was probably a typo.

Thirdly, you invoque the squeeze theorem after line #4 tu justify that Bn->x. But this is just a consequence of the axiom of the real numbers according to which for all x,y in R, we can only have one of the 3: x<y, x=y, x>y. So if we encounter an inequality of the type y\leq x \leq y, it must be that y=x. So lim Bn = x.


But in essence, your had the right proof.

 

[quasar987]

The real rigourous proof involves the old N and \epsilon though... because that's the language to use when one talks about limits.

 

[master_coda]

It seems to me that this proof assumes that \lim_{n\rightarrow\infty}B_n exists.

 

[steven187]

hello guys

well I had proved it through N-E method in which was succesful, based upon your replies above, would i be right to say that my original method is not sufficiant enough to prove it since i have used the assumption that the limit of Bn exists and that i have used the sqeeze theorem while trying to prove the sqeeze theorem, which is i think they call a fallacy,

steven

 

[quasar987]

I already pointed out that what you did after line #4 is not used the squeeze theorem but simply the axioms of the real numbers.

But I think master coda has a good point.. and in an exam that proof wouldn't be worth many points imo.