Since $(x_n)$ does not converge to $L$ there is $\epsilon >0$ such that
$|x_n-L|> \epsilon$ for infinitely many n.
Now, its your turn.
Since $(x_n)$ does not converge to $L$ there is $\epsilon >0$ such that
$|x_n-L|> \epsilon$ for infinitely many n.
Now, its your turn.