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Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits and Derivatives3 Marks
Question
Evaluate:
$\lim\limits_{\text{h} \rightarrow 0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}}$ 
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Answer

Given that $\lim\limits_{\text{h} \rightarrow 0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}\big[\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big]}\times\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}$ 
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{x}+\text{h}-\text{x}}{\text{h}\big[\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big] }$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{\text{h}\big[\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big] }$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{1}}{\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}} $
Talking the limits, We have
$\frac{1}{\sqrt{\text{x}}+\sqrt{\text{x}}}=\frac{1}{2\sqrt{\text{x}}}$
Hence, the answer is $\frac{1}{2\sqrt{\text{x}}}.$

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