If lim _{x rightarrow 0} frac{sin ^{-1} x - t | vedclass

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

Question

(D) We use the Taylor series expansions for $\sin^{-1} x$ and $	an^{-1} x$ near $x = 0$: $\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$ $	an^{-1} x = x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We use the Taylor series expansions for $\sin^{-1} x$ and $	an^{-1} x$ near $x = 0$: $\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$ $	an^{-1} x = x - \frac{x^3}{3} + O(x^5)$ Substituting these into the limit expression: $L = \lim_{x ightarrow 0} \frac{(x + \frac{x^3}{6}) - (x - \frac{x^3}{3})}{3x^3}$ $L = \lim_{x ightarrow 0} \frac{\frac{x^3}{6} + \frac{x^3}{3}}{3x^3} = \lim_{x ightarrow 0} \frac{\frac{1}{6} + \frac{1}{3}}{3} = \frac{\frac{1+2}{6}}{3} = \frac{3/6}{3} = \frac{1/2}{3} = \frac{1}{6}$ Given $L = 1/6$,we calculate $6L + 1$: $6(1/6) + 1 = 1 + 1 = 2$

If $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x - \tan ^{-1} x}{3 x^{3}}$ is equal to $L$,then the value of $(6L + 1)$ is

Similar Questions

Find the principal value of $\tan ^{-1}(-\sqrt{3})$

If $\cos ^{-1}\left(\frac{1}{2}\right)=\cot \left(\cos ^{-1} x\right)$,then the value of $x$ is

The value of $\tan ^{-1}(-\sqrt{3})-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)$ is

If $y = \cos^{-1}(\cos(|x| - f(x)))$,where $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \end{cases}$,then $\left. \frac{dy}{dx} \right|_{x = \frac{5\pi}{4}}$ is

$\cos (\tan ^{ - 1}x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module