According to question
\(=\lim\limits_{h\to 0} \frac{-h}{h\sqrt{x(x + h) }(\sqrt x + \sqrt{x + h})}\)
\(=\lim\limits_{h\to 0} \frac{-1}{\sqrt{x(x + h) }(\sqrt x + \sqrt{x + h})}\)
\(=\frac{-1}{\sqrt{x(x + 0) }(\sqrt x + \sqrt{x + 0})}\)
\(= \frac{-1}{\sqrt {x^2} (\sqrt x + \sqrt x)}\)
\(= \frac{-1}{x \times 2\sqrt x}\) \((\because \sqrt{x^2 } =x)\)
\(= \frac{-1}{2x\sqrt x}\)
Alternative:
\(\lim\limits_{h\to 0} \frac 1h\left\{\frac1{\sqrt x + h} - \frac1{\sqrt x}\right)\)
\(= \lim\limits_{h \to 0} \frac{\sqrt x - \sqrt{x + h}}{h\sqrt x\sqrt{x + h}}\) (By taking LCM)
\(= \lim\limits_{h \to 0} \frac{(\sqrt x-\sqrt{x +h})}{h\sqrt x \sqrt{x + h}} \times \frac{\sqrt x + \sqrt {x+h}}{\sqrt x+ \sqrt {x + h}}\)
\(= \lim\limits_{h\to 0} \frac{x - (x+ h)}{h\sqrt x \sqrt{x +h}(\sqrt x + \sqrt{x +h})}\) \((\because (a-b)(a+b)= a^2-b^2)\)
\(= \lim\limits_{h\to 0}\frac{-h}{h\sqrt x \sqrt{x+h}(\sqrt x + \sqrt {x +h})}\)
\(= \frac{-1}{\sqrt x \sqrt{x +0}(\sqrt x + \sqrt{x +0})}\) (By taking LCM)
\(= \frac{-1}{2\sqrt x\sqrt{x^2}}\)
\(=\frac{-1}{2x\sqrt x}\)