નીચે બે વિધાનો આપેલા છે:વિધાન I: lim _{x righ | vedclass

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

Question

(D) For Statement $I$:Using Taylor series expansions:$	an ^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$$\log _e \sqrt{\frac{1+x}{1-x}} = \frac

Vedclass · Accepted Answer

બંને વિધાન $I$ અને વિધાન $II$ સાચા છે

Vedclass · Answer

(D) For Statement $I$:Using Taylor series expansions:$	an ^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$$\log _e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2} [\ln(1+x) - \ln(1-x)] = \frac{1}{2} [(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}) - (-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5})] = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$Substituting these into the limit:$\lim _{x ightarrow 0} \frac{(x - \frac{x^3}{3} + \frac{x^5}{5}) + (x + \frac{x^3}{3} + \frac{x^5}{5}) - 2x}{x^5} = \lim _{x ightarrow 0} \frac{2x + \frac{2x^5}{5} - 2x}{x^5} = \frac{2}{5}$.Thus,Statement $I$ is true.For Statement $II$:Let $L = \lim _{x ightarrow 1} x^{\frac{2}{1-x}}$. This is a $1^\infty$ form.$L = e^{\lim _{x ightarrow 1} (x-1) \cdot \frac{2}{1-x}} = e^{\lim _{x ightarrow 1} -2} = e^{-2} = \frac{1}{e^2}$.Thus,Statement $II$ is true.

Similar Questions

જો $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$ હોય,તો $\frac{\log _e \alpha}{1+\log _e \alpha}$ ની કિંમત શોધો:

$\lim _{x \rightarrow 0} \frac{|x|}{|x|+x^2} = $

જો $a > 0$ અને $n \in R$ હોય,તો $\lim_{x \rightarrow a} x^n = \dots$

$\mathop {\lim }\limits_{x \to \infty } \left( {\left| {{x^2}} \right| + x} \right)\log \left( {x{{\cot }^{ - 1}}x} \right)$ ની કિંમત શોધો.

જો $x > 2$ માટે $g(x) = \frac{x}{[x]}$ હોય,તો $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module