Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The "Power Rule" for null sequences states that

If a null sequence of non-negative terms is raised to a positive power, the resulting sequence is also a null sequence.

Ok, can this rule be generalised to the following?

If a sequence of non-negative terms that converges to the limit $l$ is raised to a positive power $p$, the resulting sequence converges to the limit $l^p$.

share|improve this question
2  
The function $f(x)=x^p$ is continuous, so ... –  Brian M. Scott May 11 '12 at 21:01
    
If $\langle x_n:n\in\Bbb N\rangle\to x$, and $f$ is continuous, what can you say about $\langle f(x_n):n\in\Bbb N\rangle$? –  Brian M. Scott May 11 '12 at 21:31
    
Hang on, and I'll write up an answer. –  Brian M. Scott May 11 '12 at 21:39
    
The motivation for my question is that my textbook gives the following Combination Rules for the limits of convergent sequences: sum, multiple, product, quotient, reciprocal rule. And I was wondering why the power rule for null sequences is not also extended here. And I see from math.stackexchange.com/q/83460/21813 that the power rule does apply for the limits of functions. –  Ryan May 11 '12 at 21:42
add comment

1 Answer

up vote 1 down vote accepted

Proposition: Suppose that $f:[0,\infty)\to[0,\infty)$ is continuous. Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $[0,\infty)$ converging to $a$. Then $\langle f(x_n):n\in\Bbb N\rangle\to f(a)$.

Proof: Let $\epsilon>0$; since $f$ is continuous, there is a $\delta>0$ such that $|f(x)-f(x)|<\epsilon$ whenever $|x-a|<\delta$. Since $\langle x_n:n\in\Bbb N\rangle\to a$, there is an $n_0\in\Bbb N$ such that $|x_n-a|<\delta$ whenever $n\ge n_0$. But then $|f(x_n)-f(a)|<\epsilon$ whenever $n\ge n_0$, so $\langle f(x_n):n\in\Bbb N\rangle\to f(a)$. $\dashv$

This result actually holds in much greater generality: if $X$ and $Y$ are topological spaces, $f:X\to Y$ is continuous, and $\langle x_n:n\in\Bbb N\rangle\to a$ in $X$, then $\langle f(x_n):n\in\Bbb N\rangle\to f(a)$ in $Y$; the proof is identical if $X$ and $Y$ are metric spaces and very similar even for general topological spaces. In words, continuous functions preserve convergent sequences.

Now just observe that if $p>0$, the function $f:[0,\infty)\to[0,\infty):x\mapsto x^p$ is continuous, and your generalization follows immediately.

share|improve this answer
    
What a useful theorem! Thank you. –  Ryan May 11 '12 at 22:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.