Prove that if $\lim a_n = a$, then $\lim a_n^3 = a^3$.
Prove that if $\lim a_n = a$, then $\lim a_n^3 = a^3$.
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real-analysis
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0What have you tried? You probably need to use the definition of limit. You might need two different cases, one where $a=0$ and one where $a\neq 0$. – 2017-02-16
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0All polynomial (so the monomial $x^3$) is continuous. Notice that you consider implicitely that $x$ is a real. You have then $\lim a_n=a$ implies the end. – 2017-02-16
2 Answers
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If $a_n \to a$, then it is bounded. Thus $|a_n| \le M$ for all $n$. Thus, let $K = \text{max}(M,|a|)$, then $|a_n^3 - a^3| = |a_n-a||a_n^2+ a\cdot a_n + a^2|\le 3K^2|a_n-a|< \epsilon$, when $|a_n-a| < \dfrac{\epsilon}{3K^2}$ for $n > N_0$.
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0Thanks! That helps a lot! :] – 2017-02-16
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Hint:
$$ x^3-y^3=(x-y)(x^2+xy+y^2) $$
You know that $a-a_n$ tends to $0$. So, it is sufficient to show that $a^2+aa_n+a_n^2$ is bounded. Can you see how to do that using the facts that (1) $a$ is constant and (2) $a_n$ converges?