$$\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$$
The proof that I have:
Let $\varepsilon > 0$, we must show that
$$\exists \delta >0: 0<x-1<\delta \Rightarrow \left | \frac{1}{\sqrt{x}}-1\right|<\epsilon$$
So usually, when doing these $\varepsilon,\delta$ proofs, I would write $\ldots 0<|x-1|<\delta \ldots$ , but as $x\to 1+$, $x-1$ should always be greater than $0$. Is that correct? Can I just take $|x-1|=x-1$?
We see that $$ \left | \frac{1}{\sqrt{x}}-1\right|= \left | \frac{1-\sqrt{x}}{\sqrt{x}}\right|=\left | \frac{-(\sqrt{x}-1)}{\sqrt{x}}\right|=\left | \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}\right|=\left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|\left(<\varepsilon\right)$$ We know that $0<x-1<\delta$. But we don't know the estimation for $\frac{1}{\sqrt{x}(\sqrt{x}+1)}$. Let $\delta \leq 1$, so $$ \begin{align} 0<&x-1<\delta\leq 1\\ 0<&x-1<1\\ 1<&x<2\\ 1<&\sqrt{x}<\sqrt{2}\\ \frac{1}{\sqrt{2}}<&\frac{1}{\sqrt{x}}<1 \end{align} $$
If I had $|x-1|<\delta$, I would take $\delta\leq\frac{1}{2}$ and get $$ \begin{align} &|x-1|<\frac{1}{2}\\ -\frac{1}{2}<&x-1<\frac{1}{2}\\ \frac{1}{2}<&x<\frac{3}{2}\\ \frac{1}{\sqrt{2}}<&\sqrt{x}<\frac{\sqrt{3}}{\sqrt{2}}\\ &\vdots \end{align} $$
The estimation for $\frac{1}{\sqrt{x}}$ is known, let's do the same for $\frac{1}{\sqrt{x}+1}$: $$ \begin{align} 1<&\sqrt{x}<\sqrt{2}\\ 2<&\sqrt{x}+1<\sqrt{2}+1\\ \frac{1}{\sqrt{2}+1}<&\frac{1}{\sqrt{x}+1}<\frac{1}{2} \end{align} $$ So the estimation for $\frac{1}{\sqrt{x}(\sqrt{x}+1)}$ is $$ \frac{1}{\sqrt{x}(\sqrt{x}+1)}<\frac{1}{2} $$ Finally we get $$ \left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|<\frac{\delta}{2}\leq \varepsilon $$ $$\delta:=\min\{1,2\varepsilon\}$$
How to prove it? Is my approach correct? Any pointers, when doing one-sided limit proofs?