Given below are two statements:Statement I: l | 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

Both Statement $I$ and Statement $II$ are true

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

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 2} \frac{3x^{2}-x-10}{x^{2}-4}$

For each $x \in \mathbb{R}$,let $[x]$ represent the greatest integer function. Then $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$ is equal to

Evaluate the given limit: $\mathop {\lim }\limits_{x \to \pi } \left(x-\frac{22}{7}\right)$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e}}{x}$ equals

If $[x]$ is the greatest integer function,then $\lim _{x \rightarrow 3^{-}} \frac{(3-|x|+\sin |3-x|) \cos [9-3 x]}{|3-x|[3 x-9]} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module