Real Analysis - Limits and Order #2

**ajj86** · October 22nd 2008, 06:56 PM

Let (an) be a bounded (not necessarily convergent) sequence, and assume
lim bn = 0. Show that lim(an*bn) = 0. Why are we not allowed to use the Algebraic Limit Theorem to prove this.

**Moo** · October 22nd 2008, 11:42 PM

Hello,

Originally Posted by ajj86

Let (an) be a bounded (not necessarily convergent) sequence, and assume
lim bn = 0. Show that lim(an*bn) = 0. Why are we not allowed to use the Algebraic Limit Theorem to prove this.

What is the algebraic limit theorem ??

$\lim b_n=0 \Leftrightarrow \forall \epsilon > 0, ~ \exists N \in \mathbb{N}, ~ \forall n > N, ~ |b_n|< \epsilon$

$(a_n) \text{ bounded } \Leftrightarrow \exists K>0 \quad |a_n|<K \quad \forall n$ (in particular for n>N)

Multiply :
$|a_n| \times |b_n|=|a_n \times b_n|< \epsilon K$

So :
$\forall \delta= \epsilon K, ~ \exists N \in \mathbb{N}, ~ \forall n > N, ~ |a_n| \times |b_n|=|a_n \times b_n|< \delta$

This is the definition for $\lim a_n \times b_n=0$

**ajj86** · October 23rd 2008, 06:24 PM

I appreciate your answer to my question. I've been having a lot of trouble understanding real analysis. This class reminds me a lot of abstract algebra. I guess proofs just aren't my forte.

Math Help - Real Analysis - Limits and Order #2

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Real Analysis - Limits and Order #2

Thanks Moo.

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