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

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(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.

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