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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 $l^p$.

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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 that the power rule does apply for the limits of functions. – Ryan May 11 '12 at 21:42
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.

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What a useful theorem! Thank you. – Ryan May 11 '12 at 22:47

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