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Lim X → π 4 √ 2 − Cos X − Sin X ( π 4 − X ) 2

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प्रश्न

\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\] 

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उत्तर

\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
\[\text{ Dividing the numerator and the denominator by }\sqrt{2}:\]
\[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \left( \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x \right)}{\frac{1}{\sqrt{2}} \left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{1 - \left( \cos \frac{\pi}{4}\cos x + \sin \frac{\pi}{4} \sin x \right)}{\frac{1}{\sqrt{2}} \left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\left( 1 - \cos \left( \frac{\pi}{4} - x \right) \right)}{\left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \left( 2 \sin^2 \left( \frac{\frac{\pi}{4} - x}{2} \right) \right)}{\left( \frac{\pi}{4} - x \right)^2} \left[ \because 1 - \cos \theta = 2 \sin \frac{2\theta}{2} \right]\]
\[ = 2\sqrt{2} \lim_{x \to \frac{\pi}{4}} \frac{\left[ \sin^2 \left( \frac{\frac{\pi}{4} - x}{2} \right) \right]}{4 \left( \frac{\frac{\pi}{4} - x}{2} \right)^2}\]
\[ = \frac{2\sqrt{2}}{4}\]

\[ = \frac{1}{\sqrt{2}}\]

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 12 | पृष्ठ ६२

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