mathop {lim }limits_{x to 0} frac{1}{x}left[ | vedclass

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Question

(B) हम जानते हैं कि $\frac{\pi}{4} = 	an^{-1}(1)$.माना $L = \mathop {\lim }\limits_{x 	o 0} \frac{1}{x}\left[ {{{	an }^{ - 1}}\left( {\frac{{x + 1}

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(B) हम जानते हैं कि $\frac{\pi}{4} = 	an^{-1}(1)$.माना $L = \mathop {\lim }\limits_{x 	o 0} \frac{1}{x}\left[ {{{	an }^{ - 1}}\left( {\frac{{x + 1}}{{2x + 1}}} ight) - {{	an }^{ - 1}}(1)} ight]$.सूत्र $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$ का उपयोग करने पर:$L = \mathop {\lim }\limits_{x 	o 0} \frac{1}{x} 	an^{-1}\left( \frac{\frac{x+1}{2x+1} - 1}{1 + \frac{x+1}{2x+1}} ight)$$L = \mathop {\lim }\limits_{x 	o 0} \frac{1}{x} 	an^{-1}\left( \frac{x+1 - 2x - 1}{2x+1 + x+1} ight) = \mathop {\lim }\limits_{x 	o 0} \frac{1}{x} 	an^{-1}\left( \frac{-x}{3x+2} ight)$$\mathop {\lim }\limits_{	heta 	o 0} \frac{	an^{-1}(	heta)}{	heta} = 1$ का उपयोग करने पर:$L = \mathop {\lim }\limits_{x 	o 0} \left( \frac{	an^{-1}\left( \frac{-x}{3x+2} ight)}{\frac{-x}{3x+2}} ight) 	imes \left( \frac{1}{x} \cdot \frac{-x}{3x+2} ight)$$L = 1 	imes \mathop {\lim }\limits_{x 	o 0} \left( \frac{-1}{3x+2} ight) = -\frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{2x + 1}}} \right) - \frac{\pi }{4}} \right]$ का मान है

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