Question regarding the limit of $f(x)= \sqrt{x+1} - \sqrt{x}$ using the MVT

Question

I need help understanding the mean value theorem.

Let $f:\mathbb{R^+} \cup \{0\} \to \mathbb{R}$ with $f(x)= \sqrt{x+1} - \sqrt{x}$

We use $g(x) = \sqrt{x}$

Per the MVT, there exists $\xi \in (x,x+1)$:

$$g'(\xi) = \frac{g(x+1)-g(x)}{x+1-x} = \sqrt{x+1} - \sqrt{x} \stackrel{!}{=} \frac{1}{2\sqrt{\xi}} \to 0 \text{ for } x \to \infty $$

since $x < \xi < x+1$.

What I don't understand are the following points:

$1.$ How do we get to $\frac{g(x+1)-g(x)}{x+1-x}$?

$2.$ How does that lead to $\frac{1}{2\sqrt{\xi}}$?

According to Wikipedia the general formula is $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$.

I think this formula was used for $\frac{g(x+1)-g(x)}{x+1-x}$, but how do we get to $h$?

Answer 1

Let $x>0$, then $g(x)=\sqrt x$ is continuous in $[x,x+1]$ and differentiable in $(x,x+1)$ So from the mean value theorem we have that there is $\xi$ in $(\color{red}{x},\color{blue}{x+1})$ sucha that $$ g'(\xi)=\frac{\color{blue}{g(x+1)}-\color{red}{g(x)}}{\color{blue}{(x+1)}-\color{red}{x}}=\sqrt{x+1}-\sqrt x\Rightarrow f(x)=\sqrt{x+1}-\sqrt x=g'(\xi) $$ On the other hand we have to: $$ g(x)=\sqrt x\Rightarrow g'(x)=\dfrac{1}{2\sqrt x}\Rightarrow g'(\xi)=\dfrac{1}{2\sqrt \xi} $$ Therefore there is $\xi\in (x,x+1)$ such that $f(x)=\dfrac{1}{2\sqrt \xi}$. If we do $x\to\infty$, then $\xi\to\infty$ also, since $\xi\in (x,x+1)$. Hence $$ \lim_{x\to +\infty}f(x)=\lim_{x\to +\infty}(\sqrt{x+1}-\sqrt x)=\lim_{\xi\to +\infty}\dfrac{1}{2\sqrt \xi}=0. $$

Answer 2

1. $g(x+1)-g(x)=[(x+1)-x] g'(\xi)$ for some $\xi$ between $x$ and $x+1$. Divide by (x+1)-x.

2. $\frac 1 {2\sqrt {\xi}} $ is the value of $g'(\xi)$.

Answer 3

It is common to write the Mean Value Theorem as $$ \frac{g(y)-g(x)}{y-x}=g'(\xi) $$ for some $\xi$ between $x$ and $y$. This is equivalent to the formulation you have, just by taking $h=y-x$. In this case, $y=x+1$.

The derivative of $g(x)=\sqrt x$ is $g'(x)=\frac1{2\sqrt x}$, so $$ g'(\xi)=\frac1{2\sqrt \xi}. $$