lim _{x rightarrow 0^{+}} frac{x sin ^{-1}lef | vedclass

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

Question

(B) આપેલ લક્ષ: $\lim _{x ightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}ight)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight) 	an ^{-1}\le

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) આપેલ લક્ષ: $\lim _{x ightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}ight)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight) 	an ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}ight)}$.$x ightarrow 0^{+}$ હોવાથી,આપણે નીચેના ત્રિકોણમિતીય નિત્યસમનો ઉપયોગ કરીએ છીએ:$2 	an ^{-1} x = \sin ^{-1} \left(\frac{2 x}{1+x^2}ight) = \cos ^{-1} \left(\frac{1-x^2}{1+x^2}ight)$$3 	an ^{-1} x = 	an ^{-1} \left(\frac{3 x-x^3}{1-3 x^2}ight)$આ કિંમતો લક્ષમાં મૂકતા:$\lim _{x ightarrow 0^{+}} \frac{x \cdot (2 	an ^{-1} x)}{(2 	an ^{-1} x) \cdot (3 	an ^{-1} x)} = \lim _{x ightarrow 0^{+}} \frac{x}{3 	an ^{-1} x}$.પ્રમાણિત લક્ષ $\lim _{x ightarrow 0} \frac{	an ^{-1} x}{x} = 1$ નો ઉપયોગ કરતા:$\lim _{x ightarrow 0^{+}} \frac{1}{3 \left(\frac{	an ^{-1} x}{x}ight)} = \frac{1}{3 	imes 1} = \frac{1}{3}$.

$\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}$ ની કિંમત શોધો.

Similar Questions

જો $y = \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right)$ હોય,તો $\frac{dy}{dx} = $

$\sum_{i=0}^2 \cot ^{-1}\{-(i+1)\}=$ . . . . . . .

$\cos^{-1}\left(\frac{15}{17}\right) + 2\tan^{-1}\left(\frac{1}{5}\right) = $

કિંમત શોધો: $\tan^{-1} \left( \frac{1}{\sqrt{x^2 - 1}} \right)$

જો $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$ હોય, તો

Vedclass Test Series

Exam Paper Generator

Online Exam Module