mathop {lim }limits_{x to pi /4} frac{{sqrt 2 | vedclass

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Question

(B) ધારો કે $L = \mathop {\lim }\limits_{x 	o \pi /4} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}}$.આ સ્વરૂપ $\frac{0}{0}$ હોવાથી,આપણે $L$-Hospital નો ન

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) ધારો કે $L = \mathop {\lim }\limits_{x 	o \pi /4} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}}$.આ સ્વરૂપ $\frac{0}{0}$ હોવાથી,આપણે $L$-Hospital નો નિયમ વાપરીએ:$L = \mathop {\lim }\limits_{x 	o \pi /4} \frac{{\frac{d}{{dx}}(\sqrt 2 \cos x - 1)}}{{\frac{d}{{dx}}(\cot x - 1)}}$$L = \mathop {\lim }\limits_{x 	o \pi /4} \frac{{-\sqrt 2 \sin x}}{{-\csc^2 x}}$$L = \mathop {\lim }\limits_{x 	o \pi /4} \sqrt 2 \sin x \sin^2 x = \sqrt 2 \sin^3 x$$x = \pi /4$ મૂકતા:$L = \sqrt 2 \left( \frac{1}{{\sqrt 2 }} ight)^3 = \sqrt 2 \cdot \frac{1}{{2\sqrt 2 }} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to \pi /4} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}} = $

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