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Let $C$ be a real number such that $|C|<1$. Show $$\lim_{x\to\infty}C^{n}=0$$

What I did so far: I made some cases.
When $C=0$ : then you have the constant sequence $0,0,0,\dots$, so that converges to $0$. (Let $N=1$.)

When $0<C<1$ : then the numbers in the sequence decrease because you're getting always strictly less than 100 % of the previous number. You can never go negative from multiplying positive numbers, so the sequence is bounded and decreasing. So the limit exists and now we have to show it's $0$. To find the right $N$, I tried to consider when $|C^{n}-0|<\epsilon\iff C^{n}<\epsilon$. I didn't know how to do the algebraic manipulations, but I think the idea is that given an $\epsilon$, you have to multiply $C$ enough times until it's $<\epsilon$, and "enough times" depends on $C$ (like if $C$ is larger, you'll have to multiply more of them because the decrease is slower). But I don't know how to write a specific formula for $N$. So my question is, how do I find a formula for $N$? (Or how do I show $N$ exists without saying what it should be?) I was similarly stuck on the $-1<C<0$ case.

The book has solutions similar to what I have done so far, but I don't understand this: "We have $C^{n+1}=CC^{n}$ so by passing to the limit we get $L=CL$ which implies $L=0$." What does "passing to the limit" mean?

I'm self-learning analysis and doing practice problems in Aksoy and Khamsi's A Problem Book in Real Analysis. This is from Chapter 3, Problem 3.3.

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  • $\begingroup$ It might be slightly more convenient to prove $C^n \rightarrow \infty$ if $C >1$ first. To do this you can use the inequality $(1+x)^n \geq 1 + nx$. $\endgroup$ – Jair Taylor Jun 23 '17 at 23:51
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To find a specific formula for $N$, you can use logarithms:

$$C^n < \varepsilon \iff n > \log_C \varepsilon.$$

Note the change of direction of inequality, as logarithms with bases strictly less than $1$ are strictly decreasing functions. I don't particularly like this approach, as logarithms are objects that need to be defined in their own good time in a real analysis course, and using them now for $\varepsilon$-$N$ proofs is somewhat jumping the gun.

I prefer the book's approach. There are already two answers explaining it.

For the $C < 0$ case, I suggest considering $|C|^n$. You already know it tends to $0$, from the $C > 0$ case. It should be clear, using that fact and the $\varepsilon$-$N$ definition of a limit, that the $C < 0$ case works too.

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  • $\begingroup$ Yes, there was a previous problem about showing that if $(a_{n})$ converges to $\ell$, then $(|a_{n}|)$ converges to $|\ell|$. Thank you for your help! $\endgroup$ – anonanon444 Jun 24 '17 at 2:19
  • $\begingroup$ You're welcome! But I should point out that this is a partial converse: if $|a_n| \rightarrow 0$, then $a_n \rightarrow 0$. It doesn't work with any limit other than $0$! $\endgroup$ – Theo Bendit Jun 24 '17 at 2:36
  • $\begingroup$ Oh right, thanks again haha :) $\endgroup$ – anonanon444 Jun 24 '17 at 21:32
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First you need to show the limit exists. This is accomplished by the monotone convergence theorem. Once you know that the limit exists, you may use the algebraic manipulations. Let $L = \lim_{n \to \infty}C^n$, then $$\lim_{n \to \infty}C^n = C \cdot\lim_{n\to \infty} C^{n-1}$$ $$L = C \cdot L$$ $$L = 0$$ When a textbook says "passing to the limit", it just means to take the limit of the expression or equation.

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  • $\begingroup$ Thanks for your explanation! $\endgroup$ – anonanon444 Jun 24 '17 at 2:19
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"Passing to the limit" means taking the limit of both sides. So if you assume that the limit of $C^n$ exists, and you have this equation: $$C^{n+1}=CC^n$$ then $$\lim_{n\to\infty}C^{n+1}=C\lim_{n\to\infty}C^n$$ and if the limit exists then the sequences $C^n, C^{n+1}$ must have the same limit.

Note that you can do that only if you already know that the limit exists.

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  • $\begingroup$ I understand now, thank you! $\endgroup$ – anonanon444 Jun 24 '17 at 2:19

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