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Question

Question: \[\underset{\mathbf{h \rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\sqrt{\mathbf{x +}\mathbf{h}}\math...

lim⁡h→0x+h−xh=\underset{\mathbf{h \rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\sqrt{\mathbf{x +}\mathbf{h}}\mathbf{-}\sqrt{\mathbf{x}}}{\mathbf{h}}\mathbf{=}

A

1/2x1/2\sqrt{x}

B

1/2h1/2\sqrt{h}

C

Zero

D

None of these

Answer

1/2x1/2\sqrt{x}

Explanation

Solution

lim⁡h→0x+h−xh\lim_{h \rightarrow 0}\frac{\sqrt{x + h} - \sqrt{x}}{h} = lim⁡h→0x+h−xh×x+h+xx+h+x\lim_{h \rightarrow 0}\frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}

= lim⁡h→0(x+h)−xh(x+h+x)\lim_{h \rightarrow 0}\frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x)}} = lim⁡h→0hh(x+h+x)\lim_{h \rightarrow 0}\frac{h}{h(\sqrt{x + h} + \sqrt{x})} = 12x\frac{1}{2\sqrt{x}}

Trick : Applying ‘L’ Hospital’s rule, [Differentiating Nr and Dr with respect to h]

We get, lim⁡h→012x+h−01=12x\lim_{h \rightarrow 0}\frac{\frac{1}{2\sqrt{x + h}} - 0}{1} = \frac{1}{2\sqrt{x}}.