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Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits and Derivatives5 Marks
Question
Evaluate the following limits.
$\lim\limits_{\text{y} \rightarrow 0}\frac{(\text{x}+\text{y})\sec(\text{x}+\text{y})-\text{x}\sec\text{x}}{\text{y}}$
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Answer

$\lim\limits_{\text{y} \rightarrow 0}\frac{(\text{x}+\text{y})\sec(\text{x}+\text{y})-\text{x}\sec\text{x}}{\text{y}}$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\sec(\text{x}+\text{y})+\text{y}\sec(\text{x}+\text{y})-\text{x}\sec\text{x}}{\text{y}}$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\Big[\text{x}\sec(\text{x}+\text{y})-\text{x}\sec\text{x}\big]}{\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\frac{\text{y}\sec(\text{x}+\text{y})}{\text{y}}$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\big[\sec(\text{x}+\text{y})-\sec\text{x}\big]}{\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\big[\frac{1}{\cos(\text{x}+\text{y})}-\frac{1}{\cos\text{x}}\big]}{\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\text{x}\bigg[\frac{\cos\text{x}-\cos(\text{x}+\text{y})}{\text{y}\cdot\cos(\text{x}+\text{y})\cdot\cos\text{x}}\bigg]+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\bigg[-2\sin\Big(\frac{\text{x}+\text{x}+\text{y}}{2}\Big)\cdot\sin\Big(\frac{\text{x}-\text{x}-\text{y}}{2}\Big)\bigg]}{\text{y}\cos(\text{x}+\text{y})\cdot\text{x}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\Big[-2\sin\big(\text{x}+\frac{\text{y}}{2}\Big)\cdot\Big(\frac{-\text{y}}{2}\Big)\Big]}{\cos(\text{x}+\text{y})\cdot\cos\text{x}\cdot\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
Taking the limits we have
$=\text{x}\Big[\sin\text{x}\cdot\frac{1}{\cos\text{x}\cdot\cos\text{x}}\Big]+\sec\text{x}$
$=\text{x}\sec\text{x}\tan\text{x}+\sec\text{x}$
$=\sec\text{x}(\text{x}\tan\text{x}+1)$
Hence, the required answer is $\sec\text{x}(\text{x}\tan\text{x}+1).$

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