Prove the limit $\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$, using epsilon-delta definition. $$\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$$
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?


Coming from the right, you just have to show that $1-\frac{1}{\sqrt x} \longrightarrow$ as $x \longrightarrow 1$.  To show the limit, you could define $y=x-1$.
Then $$1-\frac{1}{\sqrt x} = \frac{\sqrt{1+y} - 1}{\sqrt{1+y}}$$.  
Note that when $y < \delta$
$$\frac{\sqrt{1+\delta} - 1}{\sqrt{1+\delta}} \leq \sqrt{1+\delta} - 1 \leq 1+\frac{\delta}{2} - 1 = \frac{\delta}{2}$$ where the last inequality is by noting that the concave square root function lies below its linear approximation at $x=1$. So, set $ \delta = 2\epsilon $ to meet the epsilon-delta definition.
