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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\sin\text{x}}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\sin\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}(1+\cos\text{x}-1)^{\frac{1}{\sin\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}(1-(1-\cos\text{x}))^{\frac{1}{\sin\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(1-2\sin^2\Big(\frac{\text{x}}{2}\Big)\Big)^{\frac{1}{\sin\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\Big(-2\sin^2\Big(\frac{\text{x}}{2}\Big)\Big)\times\Big(\frac{1}{\sin\text{x}}\Big)}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{-2\sin^2\big(\frac{\text{x}}{2}\big)}{\sin\text{x}}\Bigg)}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\begin{pmatrix}\frac{-2\sin^2\big(\frac{\text{x}}{2}\big)}{2\sin\big(\frac{\text{x}}{2}\big)\cos\big(\frac{\text{x}}{2}\big)}\end{pmatrix}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}-\tan\text{x}}$
$=\ \text{e}^0$
$=\ 1$

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