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Lim X → π / 4 √ 2 Cos X − 1 Cot X − 1 is Equal to

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प्रश्न

\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to

विकल्प

  • \[\frac{1}{\sqrt{2}}\] 

  • \[\frac{1}{2}\] 

  • \[\frac{1}{2\sqrt{2}}\] 

MCQ
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उत्तर

1/2 

\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}\]
\[\text{ Rationalising the numerator, we get }: \]
\[ = \lim_{x \to \frac{\pi}{4}} \left( \frac{\sqrt{2} \cos x - 1}{\cot x - 1} \right) \times \left( \frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x + 1} \right)\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( 2 \cos^2 x - 1 \right)}{\left( \cos x - \sin x \right)} \times \frac{\sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( \cos^2 x - \sin^2 x \right)}{\left( \cos x - \sin x \right)} \times \frac{\sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( \cos x + \sin x \right) \sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \frac{\left( \cos \frac{\pi}{4} + \sin \frac{\pi}{4} \right) \sin \frac{\pi}{4}}{\left( \sqrt{2}\cos \frac{\pi}{4} + 1 \right)}\]
\[ = \frac{\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)\left( \frac{1}{\sqrt{2}} \right)}{\sqrt{2} . \frac{1}{\sqrt{2}} + 1}\]
\[ = \frac{\left( \frac{2}{\sqrt{2}} \right) \times \frac{1}{\sqrt{2}}}{2}\]
\[ = \frac{1}{2}\]

The correct answer is B.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ७८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 12 | पृष्ठ ७८

वीडियो ट्यूटोरियलVIEW ALL [1]

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