Why does $x_n^2$ converge to $x^2$ if $x_n$ converges to $x$?
By Sophia Vance
Let $x_n$ be a convergent sequence converging to $x$
Then claim $x_n^2$ converges to $x^2$
I wish to use the definition to show this is the case.
Recall $x_n \to x$ iff $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $\forall n \geq N \Rightarrow |x_n - x| < \epsilon$
Then $|x_n^2 - x^2| = |x_n - x||x_n + x|$
But what is $|x_n + x|$?
$-\epsilon< x_n - x < \epsilon$ $\Rightarrow -\epsilon + 2x < x_n + x < \epsilon + 2x$
Then $|x_n^2 - x^2| = |x_n - x||x_n + x| < \epsilon(\epsilon+2x)$
But I wish to show that $|x_n^2 - x^2| < \epsilon$. How to deal with the $2x$ term?
$\endgroup$ 11 Answer
$\begingroup$Note that $x$ is fixed. Choose some $\epsilon>0.$ There exist some $\epsilon_0>0$, such that $\epsilon_0(\epsilon_0+2x)<\epsilon$
Since $x_n$ converges to $x$, there exist some $N>0$ such that for all $n>N$, we have $$|x_n-x|<\epsilon_0$$ and so, $$|x_n^2-x^2|<\epsilon_0(\epsilon_0+2x)<\epsilon.$$ This proves your statement.
$\endgroup$ 5
