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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\text{x}-\cos\text{a}}{\text{x}-\text{a}}$
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Answer

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

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