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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$
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Answer

$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$As $\text{x}\rightarrow\frac{\pi}{4},\frac\pi4-\text{x}\rightarrow0,$ let $\frac{\pi}{4}-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\cos\big(\frac\pi4-\text{y}\big)-\sin\big(\frac\pi4-\text{y}\big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\big[\cos\frac\pi4\cos\text{y}+\sin\frac{\pi}{4}\sin\text{y}+\sin\frac\pi4\cos\text{y}-\cos\frac\pi4\sin\text{y}\big]}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\Big(\frac{\cos\text{y}}{\sqrt{2}}+\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}\Big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{2\cos\text{y}}{\sqrt{2}}}{\text{y}^2}=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\sqrt{2}\cos\text{y}}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{(1-\cos\text{y})}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\frac{\text{y}^2}{4}}\times\frac14$
$=\sqrt{2}\times2\times\frac14\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\bigg)^2$
$=\sqrt{2}\times2\times\frac14\times1$
$=\frac{1}{\sqrt{2}}$

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