<s>[INST] 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
$\begin{align}\displaystyle\left\lvert \sqrt{1- \frac{1}{n}} -1 \right\rvert 
&= \left\lvert \frac{\sqrt{n-1}}{\sqrt{n}} -1 \right\rvert\\\\
&= \left\lvert \frac{\sqrt{n-1} - \sqrt{n}}{\sqrt{n}} \right\rvert\\\\
&= \left\lvert \frac{(\sqrt{n-1} - \sqrt{n})(\sqrt{n-1} + \sqrt{n})}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{(n-1)-n}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{-1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \frac{1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})}\\\\
& < \frac{1}{\sqrt{n}(\sqrt{n})}= \frac{1}{n} \leq \frac{1}{N} < \varepsilon\end{align}$
 [/INST] As the function values are smaller than $1$, we have to solve the inequation
$$1-\sqrt{1-\frac1n}<\epsilon$$ (we can assume $\epsilon\le1$ because for larger $\epsilon$ the inequation is always true), or
$$1-\epsilon<\sqrt{1-\frac1n},$$
$$(1-\epsilon)^2<1-\frac1n,$$
$$\frac1n<(2-\epsilon)\epsilon,$$
$$n>\frac1{(2-\epsilon)\epsilon}\ge\frac1{2\epsilon}$$
and this gives us a lower bound on $N$.
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