લક્ષ lim _{theta rightarrow 0} frac{tan (pi c | vedclass

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Question

(A) આપેલ લક્ષ $L = \lim _{	heta ightarrow 0} \frac{	an (\pi \cos ^{2} 	heta)}{\sin (2 \pi \sin ^{2} 	heta)}$ છે.નિત્યસમ $\cos^2 	heta = 1 - \si

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) આપેલ લક્ષ $L = \lim _{	heta ightarrow 0} \frac{	an (\pi \cos ^{2} 	heta)}{\sin (2 \pi \sin ^{2} 	heta)}$ છે.નિત્યસમ $\cos^2 	heta = 1 - \sin^2 	heta$ નો ઉપયોગ કરતા,$	an (\pi \cos^2 	heta) = 	an (\pi - \pi \sin^2 	heta)$ મળે.$	an (\pi - x) = -	an x$ હોવાથી,$	an (\pi - \pi \sin^2 	heta) = -	an (\pi \sin^2 	heta)$ થાય.આ કિંમત લક્ષમાં મૂકતા,$L = \lim _{	heta ightarrow 0} \frac{-	an (\pi \sin^2 	heta)}{\sin (2 \pi \sin^2 	heta)}$.જ્યારે $	heta ightarrow 0$,ત્યારે $\sin^2 	heta ightarrow 0$. ધારો કે $u = \pi \sin^2 	heta$,તો $	heta ightarrow 0$ માટે $u ightarrow 0$.પદાવલિ $\lim _{u ightarrow 0} \frac{-	an u}{\sin (2u)}$ બને છે.પ્રમાણિત લક્ષ $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$ અને $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$ નો ઉપયોગ કરતા:$L = \lim _{u}$ ${ightarrow 0} -\left( \frac{	an u}{u} ight) \left( \frac{2u}{\sin (2u)} ight) 	imes \frac{1}{2} = -1 	imes 1 	imes \frac{1}{2} = -\frac{1}{2}$.

લક્ષ $\lim _{\theta \rightarrow 0} \frac{\tan (\pi \cos ^{2} \theta)}{\sin (2 \pi \sin ^{2} \theta)}$ ની કિંમત શોધો :

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