How to prove If $\lim\limits_{n\to \infty}{a_{n}}^3 = a^{3}, \lim\limits_{n\to \infty} a_{n} = a$

Question

I do not know if $\lim\limits_{n\to \infty}{a_{n}}^3 = a^{3}$ , $\lim\limits_{n\to \infty} a_{n} = a$

Can you help me?

Answer 1

Hint. One may exploit $$ a_n^3-a^3=(a_n-a)(a_n^2+aa_n+a^2). $$

Answer 2

Since the function $c\colon\mathbb{R}\longrightarrow\mathbb{R}$ defined by $c(x)=\sqrt[3]x$ is continuous,$$\lim_{n\to\infty}{a_n}^3=a^3\implies\lim_{n\to\infty}{a_n}=a.$$

Answer 3

As $\lim(a_n)^3=a^3$ by definition of limit there exists ε>0 such that ${\vert (a_n)^3-a^3\vert}$<ε =>${\vert ((a_n)-a)((a_n)^2+a_na+a^2)\vert}<ε$ => ${\vert {a_n-a}\vert}{\vert (a_n)^2+a_na+a^2\vert}<ε$ so ${\vert {a_n-a}\vert}$≤${\vert {a_n-a}\vert}$ ${ {\vert {(a_n)^2+a_na+a^2}\vert}}$ <ε so ${\vert a_n-a\vert}<ε$ so ${lima_n}=a$

Answer 4

I am surprised that no-one talked about composition of limits...

Let $g:x\mapsto \sqrt[3]{x}$. This maps is continuous in $a^3$, so you can compose limits and $\displaystyle g(\lim_{n\to +\infty}a_n^3)=\lim_{n\to \infty}(g(a_n^3))$, that is $g(a^3)=a=\displaystyle\lim_{n\to \infty}a_n$.