# 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?

• We don't let $\delta \leq 1$ we suppose $0<x-1<1$ ! – Nosrati Sep 23 '17 at 10:36
• @MyGlasses. How can we suppose $0 < x-1 < 1$ without taking $\delta \leq 1$? – md2perpe Sep 23 '17 at 10:58
• $\delta$ is arbitrary and we choose $x$ from $0<x-1<\delta$ and $0<x-1<1$. If $\delta$ is less than $1$ so we don't need let $\delta=\min\{.,.\}$ at the end. – Nosrati Sep 23 '17 at 11:04
• @MyGlasses. No, $\delta$ is not arbitrary; $\epsilon$ is arbitrary. And we do not choose $x$ from $0 < x-1 < \delta$. Given $\epsilon>0$ we shall select $\delta>0$ such that for all $x$ with $0 < x-1 < \delta$ we have $\left| \frac{1}{\sqrt x} - 1 \right| < \epsilon.$ – md2perpe Sep 23 '17 at 11:13
• @md2perpe Indeed we have two separate neibourhood $1<x<\delta+1$ and $1<x<2$, and choose $x$ from one of them. For $x$ lies in both of them we let $\delta=\min\{.,.\}$! – Nosrati Sep 23 '17 at 11:15

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} $$\left | \frac{1}{\sqrt{x}}-1\right|= \left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|<\dfrac{|x-1|}{2}<\dfrac{\delta}{2}<\varepsilon$$ $$\delta:=\min\{1,2\varepsilon\}$$
• (+1). I like your proof a bit better than mine because it doesn't require knowledge of the shape of the $\sqrt$ function and implicitly, that it's differentiable, etc. Btw, the question is a bit strange in the first place in asking to show a limit because the function is so well behaved anyway at $x=1$ (continuous and differentiable). – Mathemagical Sep 24 '17 at 2:17
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.