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Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits and Derivatives5 Marks
Question
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^{3}}$ 
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Answer

Given that $\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^{3}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-2\sin\text{x}\cos\text{x}}{\text{x}^{3}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-(1-\cos\text{x})}{\text{x}^{3}}$
 $=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}}{\text{x}}\Big(\frac{1-\cos\text{x}}{\text{x}^{2}}\Big)$
$=\lim\limits_{\text{x} \rightarrow 0}2\Big(\frac{\sin\text{x}}{\text{x}}\Big)\bigg(\frac{\frac{2\sin^{2}\text{x}}{2}}{\text{x}^{2}}\bigg)$
$=\lim\limits_{\text{x} \rightarrow 0}2\Big(\frac{\sin\text{x}}{\text{x}}\Big)\Bigg(2\frac{\sin^{2}\frac{\text{x}}{2}}{\frac{\text{x}^{4}}{4}}\times\frac{1}{4}\Bigg)$
 $=\lim\limits_{\text{x} \rightarrow 0}2\Big(\frac{\sin\text{x}}{\text{x}}\Big)\Bigg(2\frac{\sin^{2}\frac{\text{x}}{2}}{\frac{\text{x}^{4}}{4}}\Bigg)^{2}.\frac{1}{4}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{4}{4}\Big(\frac{\sin\text{x}}{\text{x}}\Big)\lim\limits_{\frac{\text{x}}{2} \rightarrow 0}\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)$
$1.1.(1)^{2}=1$
Hence, the required answer is 1.

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