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Limits and Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits and Derivatives5 Marks
Question
Evaluate the following limits.
$\lim\limits_{\text{x} \rightarrow 0}\frac{(\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\cdot\text{x}}{\cos2\beta\text{x}-\cos2\alpha\text{x}}$ 
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Answer

Given$\lim\limits_{\text{x} \rightarrow 0}\frac{(\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\cdot\text{x}}{\cos2\beta\text{x}-\cos2\alpha\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\big[2\sin\alpha\text{x}\cdot\cos\beta\text{x}+\sin2\alpha\cdot\text{x}\big]\cdot\text{x}}{2\sin(\alpha+\beta)\text{x}\cdot\sin(\text{x}-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\big[2\sin\alpha\text{x}\cdot\cos\beta\text{x}+2\sin\alpha\text{x}\cdot\cos\alpha\text{x}\big]\cdot\text{x}}{2\sin(\alpha+\beta)\text{x}\cdot\sin(\text{x}-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\alpha\text{x}(\cos\beta\text{x}+\cos\alpha\text{x})\cdot\text{x}}{2\sin(\alpha+\beta)\text{x}\cdot\sin(\alpha-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\alpha\text{x}\Big[2\cos\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}\Big]\cdot\text{x}}{\sin(\alpha+\beta)\text{x}\cdot\sin(\alpha-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\alpha\text{x}\Big[2\cos\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}\Big]\cdot\text{x}}{2\sin\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\alpha\text{x}\cdot\text{x}}{2\sin\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{1}{2}\frac{\frac{\sin\alpha\text{x}}{\alpha\text{x}}\cdot(\alpha\text{x})\cdot\text{x}}{\Bigg[\frac{\frac{\sin\alpha+\beta}{2}\text{x}}{\frac{\alpha+\beta}{2}\text{x}}\times\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\Bigg]\Bigg[\frac{\sin\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}{\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}\times\frac{(\alpha-\beta)}{2}\text{x}\Bigg]}$
$=\frac{1}{2}\cdot\frac{\alpha\text{x}^{2}}{\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}$
$=\frac{1}{2}\begin{bmatrix}\frac{\alpha}{\Big(\frac{\alpha+\beta}{2}\Big)\Big(\frac{\alpha-\beta}{2}\Big)} \end{bmatrix}$
$=\frac{1}{2}\cdot\frac{4\alpha}{\alpha^{2}-\beta^{2}}$
$=\frac{2\alpha}{\alpha^{2}-\beta^{2}}$
Hence, the required amswer is $\frac{2\alpha}{\alpha^{2}-\beta^{2}}.$

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