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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{a}+\text{x}+\text{a}-\text{x}}{2}\big)\cos\big(\frac{\text{a}+\text{x}-\text{a}+\text{x}}{2}\big)-\sin\text{a}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}(\cos\text{x}-1)}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin\text{a}(1-\cos\text{x})}{\text{x}\sin\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}\big)\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{\tan\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\times\frac12$
$=-2\sin\text{a}\times1\times\frac12$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=-\sin\text{a}$

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