नीचे दो कथन दिए गए हैं:कथन I: lim _{x rightar | 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

दोनों कथन $I$ और कथन $II$ सत्य हैं

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

यदि $f(x) = \begin{cases} x & \text{यदि } x < 0 \\ 1 & \text{यदि } x = 0 \\ x^2 & \text{यदि } x > 0 \end{cases}$ है,तो $\mathop {\lim }\limits_{x \to 0} f(x) = $

$\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}=$

$\lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right) = $

$\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1} = $

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+a x+b}-x\right]$ जहाँ $a < 0 < b$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module