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

Question

$$\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$$ 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?

Answer 1

Corrections:

We assume that $x$ is chose from $0<x-1<1$: $$ \begin{align} 0<&x-1<1\\ 1<&x<2\\ 1<&\sqrt{x}<\sqrt{2}\\ \frac{1}{\sqrt{2}}<&\frac{1}{\sqrt{x}}<1\\ 2<&\sqrt{x}+1<\sqrt{2}+1\\ \frac{1}{\sqrt{2}+1}<&\frac{1}{\sqrt{x}+1}<\frac{1}{2} \end{align} $$ $$\left | \frac{1}{\sqrt{x}}-1\right|= \left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|<\dfrac{|x-1|}{2}<\dfrac{\delta}{2}<\varepsilon$$ $$\delta:=\min\{1,2\varepsilon\}$$

Answer 2

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.