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Prove that: If $|x|<1$, then $\lim_{n\to\infty}x^n=0$.

My thought:

Since it's a theory so I'm not able to solve it mathematically also I'm too confused about "infinity algebra". I have just memorized "infinity algebra results."

So, according to me, infinity is a very large number, therefore if any number is raised to power infinity then ultimately we'll get infinity.

Answer:It's a simple algebra if you'll increase power of $\vert x \vert<1$ then it'll keep decreasing the value hence you'll get $0$ too simple.

Edit:Thanku dear teachers for helping.I know it was too simple but sometimes we get struck on silly doubts like this.

Mr.HiggsBoson
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Hercules
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    If you don't want the question voted down, then you'll need to edit your question ASAP. First, you'll need to make your question clearer. What are you trying to do here? Second, you'll need to provide some of your own thoughts and attempts in solving this problem. – Theo Bendit Jun 29 '19 at 15:32
  • sir i have no idea.It's a theory question.I'm on summer vacation means no school maths teacher.Please take back your down vote or they'll ban me.Please i'm begging you this is the only way to get solve my maths doubt.Please – Hercules Jun 29 '19 at 15:35
  • please explain why its answer is 0 when |x|<1 – Hercules Jun 29 '19 at 15:36
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    A simpler relate question: do you know why $\lim_{n\to\infty}\left(\frac{1}{2}\right)^n=0$. –  Jun 29 '19 at 15:36
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    I haven't voted one way or another. I plan to up-vote your question if you can get it up to standards. Read How to ask a good question on meta. If you edit your question we can reverse its closure, and vote it up. – Theo Bendit Jun 29 '19 at 15:39
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    What kinds of results about limits do you know? Can you do any special cases? Could you prove, say, the easiest case, $x = 0$? – Theo Bendit Jun 29 '19 at 15:42
  • @Jack sir yes sir my question is similar to that.Can you please share link of that questing.I tried to find it on Google and in search box but they're not showing required result – Hercules Jun 29 '19 at 15:42
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    @RafaelNadal I really don't think anyone on this site is purposely trying to downvote you and get you blocked. Your question is pretty clear, so that's a good start. The next thing to asking a good question would be to tell us what you have tried so far. Did you for example, atleast try using a calculator? For example did you try to calculate $(0.1)^{10}, (0.1)^{20}, (0.1)^{50}$ etc? Basically, tell us what you have tried so far, and any ideas/ patterns you might have observed when trying to solve it. Next, also tell us what you know about limits. What kinds of limit rules do you know etc. – peek-a-boo Jun 29 '19 at 15:43
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    Telling us what you have tried, any ideas you have, and telling us what you already know allows us to gauge your understanding and write an answer at a suitable level. (Of course, follow the link above on how to ask a good question too) – peek-a-boo Jun 29 '19 at 15:44
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    @RafaelNadal: I do not understand your response. Again, can you prove $\lim_{n\to\infty}(1/2)^n=0$? –  Jun 29 '19 at 15:51
  • @peek-a-boo i know about all properties of limits and i can solve almost all question but i just got stuck in this question bcoz it's a theoretical because i am very confused like if i do x^n and n is infinity mean too big than how can it give 0 as result – Hercules Jun 29 '19 at 15:57
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    @RafaelNadal you said "infinity is a very large number, therefore if any number is raised to infinity then we'll ultimately get infinity." This is a very false statement; for it to be true, you need to assume that the number is positive and bigger than $1$. Once again, I suggest you use a calculator to calculate powers of $0.1$. You'll see that if $n$ is very big, then $(0.1)^n$ is EXTREMELY small. I suggest that before you try proving this question, you convince yourself using a calculator why the statement should be true – peek-a-boo Jun 29 '19 at 16:03
  • Thanks sir for explaining.Sorry for wasting your time on my stupid question.Your explanation has solved all my doubts related to Infinity.Thank you sir once again. – Hercules Jun 29 '19 at 16:06
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    Possible duplicate of Prove that $|x|<1$ implies that $\lim_{n \to \infty} x^n=0$ – Martin R Jun 29 '19 at 16:34

4 Answers4

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Bernoulli's inequality is all you need.

If $0 < x < 1$ then $x = \dfrac1{1+a}$ where $a = \dfrac1{x}-1 \gt 0$.

Then $(1+a)^n \ge 1+na \gt na =n(\dfrac1{x}-1)$ so

$\begin{array}\\ x^n &=\dfrac1{(1+a)^n}\\ &\lt\dfrac1{n(\dfrac1{x}-1)}\\ &=\dfrac{x}{n(1-x)}\\ &\to 0 \text{ as } n \to \infty\\ \end{array} $

marty cohen
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we know that $|x| < 1$, so $|x^n| < 1$ for every $n \in \Bbb{N}$. This means that the sequnece $a_n = x^n$ is bounded, its sufficient to prove that $a_n$ is convergent and then we can find the desires limit. Without losing generality we can suppose that $x > 0$, for $x = 0$ the statement is true, then its easy to show that: $$x^{n+1} \le x \cdot x^n < x^n \quad \Longrightarrow \quad a_{n+1} < a_n, \forall n \in \Bbb{N}$$ This means that $a_n$ is a monotone sequence and then $a_n$ is convergent and has limit $\ell$, and then: $$\lim_{n\rightarrow \infty} a_n = \ell = \lim_{n\rightarrow \infty} a_{n+1} = x \lim_{n\rightarrow \infty} a_n, \quad \Longrightarrow \quad \ell = x \cdot \ell, \quad \Longrightarrow \quad \ell = 0 $$ For $x < 0$, we can choose two distint subsequnce of $a_n$ for both even and odd power of $x$, then both of sequnces are convergent and prove is complete.

Lisbeth
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For $$-1<x<1$$ you can define $$x'=\frac{1}{x}$$ if $$x\neq 0$$

Dr. Sonnhard Graubner
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If $$-1<x<1$$ and you raise it to any power $n > 1 $ the value of it decreases.

Eg say$ x = 0.1 , x ^2 = 0.01, x^3 = 0.001$ or $x ^n = 0.1 ^ n = \frac{1}{10^n}$ and as $ n$ keeps getting bigger , the denominator approaches $\infty $ and the numerator approaches $0 $

Mr.HiggsBoson
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    Yes, you have given but an example. It still remains to be shown that in general, for all $x$ in the open interval (0,1), the sequence of powers of $x$ decreases to the limit $0$ and not to anything else. – DDS Jun 29 '19 at 17:00
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    @mlchristians Rigorous proof is all well and good , but as the OP mentioned that he is a beginner, getting an essence to why it's true is more important, atleast according to me :-) – Mr.HiggsBoson Jun 29 '19 at 17:10
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    Yes, but you did not prove it (as the question asked). Other than counter-examples, individual examples are not proofs. – DDS Jun 29 '19 at 17:41