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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sec5\text{x}-\sec3\text{x}}{\sec3\text{x}-\sec\text{x}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}-\sin\text{x}}{\sin3\text{x}-3\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\frac{\cos3\text{x}-\cos5\text{x}}{\cos3\text{x}\cos5\text{x}}}{\frac{\cos\text{x}-\cos3\text{x}}{\cos\text{x}\cos3\text{x}}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\cos3\text{x}-\cos5\text{x}}{\cos\text{x}-\cos3\text{x}}\times\frac{\cos\text{x}\cos3\text{x}}{\cos3\text{x}\cos5\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{-2\sin4\text{x}\sin(-\text{x})}{-2\sin(2\text{x})\sin(-\text{x})}\times\frac{\cos\text{x}}{\cos5\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin4\text{x}}{\sin2\text{x}}\times\frac{\cos\text{x}}{\cos5\text{x}}\Big)$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\sin4\text{x}\times\lim\limits_{\text{x}\rightarrow0}\cos\text{x}}{\lim\limits_{\text{x}\rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x}\rightarrow0}\cos5\text{x}}$
$=\frac{\Big(\lim\limits_{4\text{x}\rightarrow0}\frac{\sin4\text{x}}{4\text{x}}\times4\text{x}\Big)\big(\lim\limits_{\text{x}\rightarrow0}\cos\text{x}\big)}{\Big(\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2}\times2\text{x}\Big)\big(\lim\limits_{\text{x}\rightarrow0}\cos5\text{x}\big)}$
$=\frac{(1\times4\text{x})\times1}{1\times2\text{x}\times1}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\cos\text{x}=\cos0=1\Big]$
$=\frac{4\text{x}}{2\text{x}}$
$=2$

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