Proving that if $\lim_{n \rightarrow \infty} a_n = a$ then $\lim_{n \rightarrow \infty} a_n^2 = a^2$

Question

If we have a real sequence $\left|a_n\right|$ such that $\lim_{n \rightarrow \infty} a_n = a$, how do we prove (by an $\epsilon - N$ argument) that $\left|a_n\right|$ such that $\lim_{n \rightarrow \infty} a_{n}^{2} = a^2$?

I know you can use algebra to do to the following:

$$\left|a_n^2 - a^2\right| =\left|(a_n - a)(a_n + a)\right|$$

Where I feel like you can use the implication that $\lim_{n \rightarrow \infty} a_n = a$ to show that $(a_n-a) < a$ or something.

What's the proper way to go about this?

Answer 1

Hint

A convergent sequence is bounded. So you can also bound $\vert a+a_n\vert$.

Answer 2

Notice, we know $$\lim_{n\to \infty}a_n=a$$ Since the above limit exists as $n\to \infty$ so we can separate the limits in form of product as follows

$$\lim_{n\to \infty}a_n^2=\lim_{n\to \infty}(a_n\cdot a_n)$$$$=\lim_{n\to \infty}(a_n)\cdot \lim_{n\to \infty}(a_n)$$ $$=a\cdot a=a^2$$

Answer 3

If for every $\varepsilon > 0$ there is some $N \geq 1$ such that $|a_{n}-a| < \varepsilon$, then, taking any $\varepsilon > 0$, there is some $N \geq 1$ such that $$ |a_{n}^{2}-a^{2}| = |a_{n}-a||a_{n}+a| < 2|a|\varepsilon+\varepsilon^{2} $$ for all $n \geq N$.