Given f(x) = (sqrt(x+1)-1)/x for x ≠ 0: 1° find lim as x

Given $f(x) = \dfrac{\sqrt{x+1}-1}{x}$ for $x \neq 0$: 1° find $\displaystyle\lim_{x\to 0}f'(x)$; 2° if $f$ is extended by $f(0) = \tfrac12$, find $f'(0)$.

Step 1: Differentiate for $x \neq 0$

By the quotient rule with $u = \sqrt{x+1}-1$, $u' = \dfrac{1}{2\sqrt{x+1}}$, $v = x$, $v' = 1$: $f'(x) = \frac{\dfrac{x}{2\sqrt{x+1}} - \bigl(\sqrt{x+1}-1\bigr)}{x^{2}}$.

Clearing the inner fraction by multiplying numerator and denominator by $2\sqrt{x+1}$: \[f'(x) = \frac{x - 2(x+1) + 2\sqrt{x+1}}{2x^{2}\sqrt{x+1}} = \frac{2\sqrt{x+1}-(x+2)}{2x^{2}\sqrt{x+1}}.\]

Step 2: Rationalise to remove the indeterminacy

As $x\to 0$ both numerator and denominator vanish. Multiplying the numerator by its conjugate: \[2\sqrt{x+1}-(x+2) = \frac{4(x+1)-(x+2)^{2}}{2\sqrt{x+1}+(x+2)} = \frac{4x+4-x^{2}-4x-4}{2\sqrt{x+1}+(x+2)} = \frac{-x^{2}}{2\sqrt{x+1}+(x+2)}.\]

The factor $x^{2}$ now cancels against the $x^{2}$ in the denominator: $f'(x) = \frac{-1}{2\sqrt{x+1}\,\bigl(2\sqrt{x+1}+x+2\bigr)}$.

Step 3: Take the limit (1°)

\[ \lim_{x\to 0}f'(x) = \frac{-1}{2\cdot 1\cdot(2+2)} = -\frac{1}{8}. \]

Step 4: Compute $f'(0)$ from the definition (2°)

With the extension $f(0) = \tfrac12$: \[f'(0) = \lim_{x\to 0}\frac{f(x)-\tfrac12}{x} = \lim_{x\to 0}\frac{1}{x}\left(\frac{\sqrt{x+1}-1}{x}-\frac12\right) = \lim_{x\to 0}\frac{2\sqrt{x+1}-2-x}{2x^{2}}.\]

The numerator is the same expression as before, so the same rationalisation applies: $2\sqrt{x+1}-2-x = \frac{-x^{2}}{2\sqrt{x+1}+(2+x)}$.

\[ f'(0) = \lim_{x\to 0}\frac{-1}{2\bigl(2\sqrt{x+1}+2+x\bigr)} = \frac{-1}{2(2+2)} = -\frac18. \]

Step 5: Verification

\[\lim_{x\to 0}f(x) = \lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \lim_{x\to 0}\frac{1}{\sqrt{x+1}+1} = \frac12, \text{ so } f(0) = \tfrac12 \text{ is the right value} \;\;\checkmark\]

\[x = 10^{-4}:\quad \frac{f(x)-\tfrac12}{x} \approx -0.124994 \approx -\tfrac18 \;\;\checkmark\]

\[x = \tfrac12:\quad f'(0.5) = \frac{-1}{2\sqrt{1.5}\bigl(2\sqrt{1.5}+2.5\bigr)} \approx -0.08248 \;\;\checkmark\]