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Evaluate: 

\(\lim\limits_{x \to 0}{\frac{1}{h}}(\frac{1}{\sqrt{x + h}} - {\frac{1}{\sqrt{x}}})\)

2 Answers

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Best answer

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)

\(= \frac{-1}{2\sqrt x\sqrt{x^2}}\)

\(=\frac{-1}{2x\sqrt x}\)

+1 vote
by (51.9k points)

According to question

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