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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
$\text{Evaluate} \lim\limits_{x \rightarrow\frac{\pi}{4}} \bigg( \frac{ \sin x - \cos x}{x- \frac{\pi}{4}} \bigg)$
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Answer

$\lim\limits_{ x \rightarrow \frac{\pi}{4}} \Bigg[\frac {\sin x - \cos x} { x - \frac{\pi}{4}}\Bigg] = \lim\limits_{ x \rightarrow \frac{\pi}{4}} \sqrt{2} \Bigg[ \frac{\sin x.{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}.\cos x}}{ x- \frac{\pi}{4}}\Bigg]$

$ \lim\limits_{ x \rightarrow \frac{\pi}{4}} \sqrt{2} \Bigg[ \frac{\sin x.\cos {\frac{\pi}{{4}} - \cos x . \sin \frac{\pi}{4}}}{ x- \frac{\pi}{4}}\Bigg]$

$ \lim\limits_{ x \rightarrow \frac{\pi}{4}} \sqrt{2} \Bigg[ \frac{\sin\bigg( x - \frac{\pi}{4}\bigg){}}{ x- \frac{\pi}{4}}\Bigg] = \sqrt{2}. 1 = \sqrt{2}$

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