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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{1-\cos6\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$
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Answer

$\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{1-\cos6\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{2\sin^23\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$ $\big(1-\cos2\theta=2\sin^2\theta\big)$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{2}\sin3\text{x}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sin3\text{x}}{\big(\frac{\pi}{3}-\text{x}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin3\big(\frac{\pi}{3}+\text{h}\big)}{\frac{\pi}{3}-\big(\frac{\pi}{3}-\text{x}\big)}$ $\big(\text{Put x}=\frac{\pi}{3}+\text{h}\big)$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(\pi+3\text{h})}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\sin3\text{h}}{-\text{h}}$ $[\sin(\pi+\theta)=-\sin\theta]$
$=3\times\lim\limits_{\text{h}\rightarrow0}\frac{\sin3\text{h}}{3\text{h}}$
$3\times1$ $\Big(\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big)$
$=3$

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