In the formula f(x+h)-f(x) = h f' (x+ theta h), 0 < theta

In the formula $f(x+h)-f(x) = h\,f'\bigl(x+\theta h\bigr)$, $0\lt \theta\lt 1$, find $\theta$ as a function of $x$ and $h$ for $f(x) = \sqrt{x}$, and compute $\lim_{h\to0}\theta$.

Step 1: Write the identity out

With $f(x) = \sqrt{x}$ and $f'(x) = \dfrac{1}{2\sqrt{x}}$: \[\sqrt{x+h}-\sqrt{x} = \frac{h}{2\sqrt{x+\theta h}}. \tag{1}\]

Step 2: Rationalise the left side

\[ \sqrt{x+h}-\sqrt{x} = \frac{h}{\sqrt{x+h}+\sqrt{x}}, \]

so (1) becomes, after cancelling $h$, \[\frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{2\sqrt{x+\theta h}} \;\Longrightarrow\; \sqrt{x+\theta h} = \frac{\sqrt{x+h}+\sqrt{x}}{2}. \tag{2}\]

The mean value point is the value whose square root is the arithmetic mean of the endpoint square roots.

Step 3: Solve for $\theta$

Squaring (2): \[x+\theta h = \frac{\bigl(x+h\bigr)+2\sqrt{x\bigl(x+h\bigr)}+x}{4} = \frac{2x+h+2\sqrt{x\bigl(x+h\bigr)}}{4},\]

\[ \theta h = \frac{h-2x+2\sqrt{x\bigl(x+h\bigr)}}{4} \;\Longrightarrow\; \theta = \frac{h-2x+2\sqrt{x\bigl(x+h\bigr)}}{4h}. \]

Step 4: A convenient regrouping

\[ \theta = \frac14+\frac{\sqrt{x\bigl(x+h\bigr)}-x}{2h}. \tag{3} \]

Step 5: The limit

The second term of (3) is a difference quotient in $h$ for the function $h\mapsto\sqrt{x}\sqrt{x+h}$, whose derivative at $h = 0$ is $\dfrac{\sqrt{x}}{2\sqrt{x}} = \dfrac12$. Hence

\[ \lim_{h\to0}\frac{\sqrt{x\bigl(x+h\bigr)}-x}{2h} = \frac12\cdot\frac12 = \frac14, \qquad \lim_{h\to0}\theta = \frac14+\frac14 = \frac12. \]

Step 6: Verification

\[\text{Expanding: } \sqrt{x(x+h)} = x+\frac h2-\frac{h^{2}}{8x}+\cdots \;\Longrightarrow\; \theta = \frac12-\frac{h}{16x}+\cdots\]

\[x = 1,\ h = 0.1:\ \theta = \frac{0.1-2+2\sqrt{1.1}}{0.4} = 0.494044 \ \text{(and } \tfrac12-\tfrac{0.1}{16} = 0.49375) \;\;\checkmark\]

\[x = 1,\ h = 0.01:\ \theta = 0.4993781, \text{ closer to } \tfrac12 \;\;\checkmark\]