લક્ષની કિંમત શોધો: lim _{x rightarrow 0} frac | vedclass

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(C) ધારો કે $L = \lim _{x ightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{	an ^2 \frac{x}{2}}$.અંશનું સંમેયીકરણ કરતા:$L = \lim _{x ightarrow

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$3$

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(C) ધારો કે $L = \lim _{x ightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{	an ^2 \frac{x}{2}}$.અંશનું સંમેયીકરણ કરતા:$L = \lim _{x ightarrow 0} \frac{(1+x \sin x) - \cos x}{	an ^2 \frac{x}{2} (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.નિત્યસમ $1 - \cos x = 2 \sin ^2 \frac{x}{2}$ નો ઉપયોગ કરતા:$L = \lim _{x}$ ${ightarrow 0} \frac{x \sin x + 2 \sin ^2 \frac{x}{2}}{	an ^2 \frac{x}{2} (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.અંશ અને છેદને $x^2$ વડે ભાગતા:$L = \lim _{x}$ ${ightarrow 0} \frac{\frac{\sin x}{x} + 2 \left( \frac{\sin (x/2)}{x} ight)^2}{\left( \frac{	an (x/2)}{x} ight)^2 (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.$\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,$\lim _{x ightarrow 0} \frac{\sin (x/2)}{x} = \frac{1}{2}$,અને $\lim _{x ightarrow 0} \frac{	an (x/2)}{x} = \frac{1}{2}$ હોવાથી:$L = \frac{1 + 2(1/2)^2}{(1/2)^2 (\sqrt{1+0} + \sqrt{1})} = \frac{1 + 1/2}{(1/4)(2)} = \frac{3/2}{1/2} = 3$.

લક્ષની કિંમત શોધો: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}$

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