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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)}$
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Answer

$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)^2}$ $\Rightarrow\ \text{x}\rightarrow\frac{\pi}{2},$ then $\frac{\pi}{2}-\text{x}\rightarrow0,$ let $\frac{\pi}{2}-\text{x}=\text{y}$ $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}\rightarrow0}=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\sin\big(\frac{\pi}{2}-\text{y}\big)}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)}{\text{y}^2}\times\frac{\big(\sqrt{2-\cos\text{y}}+1\big)}{\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big({2-\cos\text{y}}-1\big)}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{{1-\cos\text{y}}}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=2\times\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)\times\frac14\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$ $=2\times1\times\frac14\times\frac{1}{1+1}=\frac14$

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