mathop {lim }limits_{x to 0} frac{{xtan 2x - | vedclass

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(D) ધારો કે $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x	an 2x - 2x	an x}}{{{{\left( {1 - \cos 2x} ight)}^2}}}$.$	an 2x = \frac{{2	an x}}{{1 -

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$\frac{1}{2}$

Vedclass · Answer

(D) ધારો કે $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x	an 2x - 2x	an x}}{{{{\left( {1 - \cos 2x} ight)}^2}}}$.$	an 2x = \frac{{2	an x}}{{1 - {{	an }^2}x}}$ અને $1 - \cos 2x = 2\sin^2 x$ નિત્યસમનો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{x\left( \frac{{2	an x}}{{1 - {{	an }^2}x}} ight) - 2x	an x}}{{{{\left( {2\sin^2 x} ight)}^2}}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{2x	an x - 2x	an x(1 - {{	an }^2}x)}}{{(1 - {{	an }^2}x) \cdot 4\sin^4 x}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{2x	an^3 x}}{{4\sin^4 x(1 - {{	an }^2}x)}}$$	an x = \frac{\sin x}{\cos x}$ હોવાથી:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{2x \cdot \frac{\sin^3 x}{\cos^3 x}}}{{4\sin^4 x(1 - {{	an }^2}x)}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{2x}}{{4\sin x \cdot \cos^3 x(1 - {{	an }^2}x)}}$$L = \mathop {\lim }\limits_{x 	o 0} \left( \frac{x}{2\sin x} ight) \cdot \mathop {\lim }\limits_{x 	o 0} \frac{1}{{\cos^3 x(1 - {{	an }^2}x)}}$$L = \frac{1}{2} \cdot \frac{1}{1(1 - 0)} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{x\tan 2x - 2x\tan x}}{{{{\left( {1 - \cos 2x} \right)}^2}}}$ ની કિંમત શોધો.

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