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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits5 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$
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Answer

$\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$
$=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}\times\frac{\sqrt{2+\cos\text{x}}+1}{{\sqrt{2+\cos\text{x}}+1}}$
$=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{(2+\cos\text{x})-1}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$
Let $\pi-\text{x}=\text{y},\text{x}\rightarrow\pi,\text{y}\rightarrow0$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos\text{x}(\pi-\text{y})}{\text{y}^2\big(\sqrt{2+\cos(\pi-\text{y})+1}\big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\text{y}^2\sqrt{2-\cos\text{y}+1}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\frac{2\sin^2\text{y}}{2}}{\text{y}^2\sqrt{2-\cos\text{y}}+1}$
$=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{1}{4}\frac{1}{\sqrt{2-\cos\text{y}}+1}$
$=2\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{2}\Big)^2\times\frac{1}{4}\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-\cos\text{0}}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-1}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{{1}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{{2}}$
$=\frac{1}{4}$

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