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Limits — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits5 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$
✓

Answer

$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$
$⇒ \text{x} → \pi,$ then $\pi-\text{x}\rightarrow0,$ let $\pi-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5+\cos(\pi-\text{y})}-2}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}\times\frac{\big(\sqrt{5-\cos\text{y}}+2\big)}{\big(\sqrt{5-\cos\text{y}}+2\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{(5-\cos\text{y}-4)}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$
$=2\times\frac14\times\frac{1}{\sqrt{4}+2}=2\times\frac14\times\frac14$
$=\frac18$

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