My book's proof that if $x = \displaystyle \lim_{k \to \infty}x_k$ exists, then it's unique: if $y$ were another limit then $$|x-y| = |(x-x_k)-(y-y_k)| \le |x-x_k|+|y-y_k| \le \frac{1}{n} \div\frac{1}{n} ~~~ (\star)$$

If $k$ is such large enough; hence $x-y = 0$ by the axiom of Archimedes.

My question is where does the very last inequality/the division in $(\star)$ come from?

Evaluate:
$\lim_{x \to 0} \frac{ln(1-x)-sin x}{1 - cos^2 x}$

@robjohn I had shown the following proposition:

Let $\rho \in C_{C}^{\infty}(\mathbb{R}^n)$ with $\int_{\mathbb{R}^n} \rho(x) dx=1, \rho \geq 0$ with $supp \rho \subset \{ |x| \leq 1 \}$.
We define $\rho_{\epsilon}(x)=\frac{1}{\epsilon^n} \rho \left( \frac{x}{\epsilon}\right), \epsilon \geq 0$.

Then $\rho_{\epsilon} \to \delta$ in $D'(\Omega)$ while $\epsilon \to 0$.

But we can't use this, since it doesn't hold that $supp n(x) \subset \{ |x| \leq 1 \}$, right?