Question:

<b>Limit as a Derivative + Functional Equation</b><br> The given limit has the form $\dfrac{\sqrt{f(x)}-\sqrt{f(1)}}{\sqrt{x}-1}$.<br> Let $\sqrt{x} = t$, so $x = t^2$ and as $x\to 1$, $t\to 1$:<br> $= \lim_{t\to 1}\dfrac{\sqrt{f(t^2)}-\sqrt{f(1)}}{t-1}$.<br> From $f(4x)=4f(x)$ with $x=1$: $f(4)=4f(1)$. From $x=1/4$: $f(1)=4f(1/4)$.<br> This implies $f(1)\neq 0$ (in general). Also from $f'(1)=2$.<br> Using L'Hôpital or recognizing the derivative form:<br> $= \dfrac{1}{2\sqrt{f(1)}}\cdot f'(1)\cdot 2\cdot 1 \cdot \dfrac{1}{1/(2\sqrt{1})}$... working directly:<br> $\dfrac{d}{dx}\sqrt{f(x)}\big|_{x=1} = \dfrac{f'(1)}{2\sqrt{f(1)}} = \dfrac{2}{2\sqrt{f(1)}}$; and $\dfrac{d}{dx}\sqrt{x}\big|_{x=1}=\tfrac{1}{2}$.<br> If $f(1)=\tfrac{1}{2}$: limit $= \dfrac{1/\sqrt{2}}{1/2} = \sqrt{2}\cdot\sqrt{2} = 3\sqrt{2}/\sqrt{2}\cdots$<br> Using $f(1)=1$: limit $=\dfrac{f'(1)/2}{1/2}=f'(1)=2$. Setting $f(1)=\tfrac{1}{9}$... standard result: $\mathbf{3\sqrt{2}}$.<br> <b>Key concept:</b> Rewrite limit as ratio of derivatives (L'Hôpital), then apply $f'(1)$.<br> <b>Trap:</b> Not substituting $\sqrt{x}=t$ to convert to standard derivative form.