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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{\sqrt{2+\cos\text{x}-1}}{(\pi-\text{x})^2}$
✓

Answer

$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{\sqrt{2+\cos\text{x}-1}}{(\pi-\text{x})^2}$
$\text{x}\rightarrow{\pi},$ then $\text{x}-{\pi}\rightarrow0,$ let $\text{x}-\pi=\text{y}$
$\Rightarrow\lim\limits_{{\text{x}\rightarrow{\pi}}}\frac{\big(\sqrt{2+\cos\text{x}}-1\big)}{(\pi-\text{x})^2}=\lim\limits_{{\text{x}-{\pi}}\rightarrow0}\frac{\sqrt{2+\cos(\text{x})}-1}{(-1)^2(\text{x}-\pi)^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2+\cos(\pi+\text{y})}-1}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)\big(\sqrt{2-\cos\text{y}}+1\big)}{\text{y}^2\big(\sqrt{2\cos\text{y}}+1\big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{(2\cos\text{y}-1)}{\big(\sqrt{2-\cos\text{y}+1}\big)\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{(1-\cos\text{y})}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$
$=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos0+1}}$
$=2\times\frac14\times\frac{1}{\sqrt{2}-1+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{1}{4}$

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