Show that $\lim \sqrt{1-\frac{1}{n}} = 1$ by definition Is my attempt correct?
Proof. Let $\varepsilon > 0$. Take $N > \frac{1}{\varepsilon}$ and let $n \geq N$. Then
$\begin{align}\displaystyle\left\lvert \sqrt{1- \frac{1}{n}} -1 \right\rvert 
&= \left\lvert \frac{\sqrt{n-1}}{\sqrt{n}} -1 \right\rvert\\\\
&= \left\lvert \frac{\sqrt{n-1} - \sqrt{n}}{\sqrt{n}} \right\rvert\\\\
&= \left\lvert \frac{(\sqrt{n-1} - \sqrt{n})(\sqrt{n-1} + \sqrt{n})}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{(n-1)-n}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{-1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \frac{1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})}\\\\
& < \frac{1}{\sqrt{n}(\sqrt{n})}= \frac{1}{n} \leq \frac{1}{N} < \varepsilon\end{align}$

One way to shorten this proof would be to use this lemma which says that for $x,y\in[0,\infty)$, we have $$\left|\sqrt{x}-\sqrt{y}\right|\leq\sqrt{\big||x|-|y|\big|}\leq\sqrt{|x-y|}$$
Then your proof would go $$\left|\sqrt{1-\frac{1}{n}}-\sqrt{1}\right|\leq\sqrt{\left|1-\frac{1}{n}-1\right|}=\frac{1}{\sqrt{n}}\leq\frac{1}{\sqrt{N}}<\varepsilon$$
for $1/N<\varepsilon^2$ and $n\geq N$.
Don't forget to mention the Archimedean Property, but otherwise, nice job :)
