mathop {lim }limits_{x to 0} {left{ {tan left | vedclass

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

Question

(C) माना $L = \mathop {\lim }\limits_{x 	o 0} {\left\{ {	an \left( {\frac{\pi }{4} + x} ight)} ight\}^{1/x}}$.$	an(A+B) = \frac{	an A + 	an B

Vedclass · Accepted Answer

$e^2$

Vedclass · Answer

(C) माना $L = \mathop {\lim }\limits_{x 	o 0} {\left\{ {	an \left( {\frac{\pi }{4} + x} ight)} ight\}^{1/x}}$.$	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ सूत्र का उपयोग करने पर,हमें $	an(\frac{\pi}{4} + x) = \frac{1 + 	an x}{1 - 	an x}$ प्राप्त होता है।अतः,$L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{1 + 	an x}}{{1 - 	an x}}} ight)^{1/x}}$.यह $1^\infty$ रूप है,इसलिए $L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} (\frac{1 + 	an x}{1 - 	an x} - 1)}$.$L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} (\frac{1 + 	an x - 1 + 	an x}{1 - 	an x})} = e^{\mathop {\lim }\limits_{x 	o 0} \frac{2 	an x}{x(1 - 	an x)}}$.चूंकि $\mathop {\lim }\limits_{x 	o 0} \frac{	an x}{x} = 1$,इसलिए $L = e^{2 	imes 1 	imes \frac{1}{1 - 0}} = e^2$.

$\mathop {\lim }\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{1/x}} = $

Similar Questions

$\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 3x}{\cos 4x - \cos 5x} = $

$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$ का मान ज्ञात कीजिए।

$\lim _{x \rightarrow 1} \frac{\tan \left(x^{2}-1\right)}{x-1}$ का मान ज्ञात कीजिए।

$\mathop {\lim }\limits_{x \to 0} x \log (\sin x) = $

$\mathop {\lim }\limits_{x \to 1} \frac{{\log x}}{{x - 1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module