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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.
Let $\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}, \\3,\text{x}=\frac{\pi}{2}\text{and } \text{f}(\text{x})=\text{f}\big(\frac{\pi}{2}\big)\end{cases}$ Find the value of k.
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Answer

Given $\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}, \\3,\text{x}=\frac{\pi}{2} \end{cases}$
$\text{L}.\text{H}.\text{L}.\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{2}}}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\cos\big(\frac{\pi}{2}+\text{h}\big)}{\pi-{2}\big(\frac{\pi}{2}-\text{h}\big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\sin\text{h}}{\pi-\pi+2\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\sin\text{h}}{2\text{h}}$
$=\frac{\text{k}}{2}.1=\frac{\text{k}}{2}$
$\text{R}.\text{H}.\text{L}.\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{2}}}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\cos\big(\frac{\pi}{2}+\text{h}\big)}{\pi-{2}\big(\frac{\pi}{2}-\text{h}\big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\sin\text{h}}{\pi-\pi-2\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{k}\sin\text{h}}{-2\text{h}}$
$=\frac{\text{k}}{2}.1=\frac{\text{k}}{2}$
we are given that $\lim\limits_{\text{h} \rightarrow\frac{\pi}{2}}\text{f}(\text{x})=3$ 
Hence, the required answer is 6.

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