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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4\text{x}-\pi)^2}$
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Answer

$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4\text{x}-\pi)^2}$
$\text{x}\rightarrow\frac{\pi}{4}$ then $\text{x}-\frac\pi4\rightarrow0,$ also $4\text{x}-\pi\rightarrow0$ let $\text{x}-\frac{\pi}{4}\rightarrow\text{y}$
$=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4)^2\big(\text{x}-\frac{\pi}{4}\big)^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\cos\big({\text{y}+\frac\pi4\big)-\sin\big({\text{y}+\frac\pi4\big)}}}{16\times\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\big(\cos\text{y}\times\frac{1}{\sqrt{2}}-\sin\text{y}\times\frac{1}{\sqrt{2}}\big)-\big(\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}\big)}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}(\cos\text{y}-\sin\text{y})-\frac{1}{\sqrt{2}}(\sin\text{y}+\cos\text{y})}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}\big[(\cos\text{y}-\sin\text{y})-(\sin\text{y}+\cos\text{y})\big]}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big(\sqrt{2}-\frac{1}{\sqrt{2}}\cos\text{y}+\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}\cos\text{y}}{\text{y}^2}$
$=\frac{\sqrt{2}}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{{2}\sin^2\frac{\text{y}}{{2}}}{\text{y}^2}$
$=\frac{\sqrt{2}}{8}\lim\limits_{\text{y}\rightarrow{0}}\Bigg(\frac{\sin\frac{\text{y}}{{2}}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14$
$=\frac{1}{16\sqrt{2}}$

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