mathop {lim }limits_{x to 1} (1 - x)tan left( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) माना $L = \mathop {\lim }\limits_{x 	o 1} (1 - x)	an \left( {\frac{{\pi x}}{2}} ight)$.$y = 1 - x$ प्रतिस्थापित करने पर,जब $x 	o 1$,तब $y 	o

Vedclass · Accepted Answer

$\frac{2}{\pi }$

Vedclass · Answer

(C) माना $L = \mathop {\lim }\limits_{x 	o 1} (1 - x)	an \left( {\frac{{\pi x}}{2}} ight)$.$y = 1 - x$ प्रतिस्थापित करने पर,जब $x 	o 1$,तब $y 	o 0$ और $x = 1 - y$.अतः $L = \mathop {\lim }\limits_{y 	o 0} y 	an \left( {\frac{{\pi (1 - y)}}{2}} ight) = \mathop {\lim }\limits_{y 	o 0} y 	an \left( {\frac{\pi }{2} - \frac{{\pi y}}{2}} ight)$.सर्वसमिका $	an \left( {\frac{\pi }{2} - 	heta } ight) = \cot 	heta$ का उपयोग करने पर:$L = \mathop {\lim }\limits_{y 	o 0} y \cot \left( {\frac{{\pi y}}{2}} ight) = \mathop {\lim }\limits_{y 	o 0} \frac{y}{	an \left( {\frac{{\pi y}}{2}} ight)}$.$\frac{\pi }{2}$ से गुणा और भाग करने पर:$L = \mathop {\lim }\limits_{y 	o 0} \frac{1}{\frac{\pi }{2}} \cdot \frac{{\frac{{\pi y}}{2}}}{{	an \left( {\frac{{\pi y}}{2}} ight)}} = \frac{2}{\pi } \cdot 1 = \frac{2}{\pi }$.

$\mathop {\lim }\limits_{x \to 1} (1 - x)\tan \left( {\frac{{\pi x}}{2}} \right) = $

Similar Questions

दिए गए सीमा (limit) का मूल्यांकन करें: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$,जहाँ $a, b, a+b \neq 0$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^{\sin x}} - 1}}{{{b^{\sin x}} - 1}} = $

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} = $

$\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$ का मान ज्ञात कीजिए।

$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\cot x}}{{1 - \cos x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module