mathop {lim }limits_{x to 0} {left{ {tan left | vedclass

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(C) ધારો કે $L = \mathop {\lim }\limits_{x 	o 0} {\left\{ {	an \left( {\frac{\pi }{4} + x} ight)} ight\}^{1/x}}$.$	an(A+B) = \frac{	an A + 	a

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$e^2$

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(C) ધારો કે $L = \mathop {\lim }\limits_{x 	o 0} {\left\{ {	an \left( {\frac{\pi }{4} + x} ight)} ight\}^{1/x}}$.$	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ સૂત્રનો ઉપયોગ કરતા,આપણને $	an(\frac{\pi}{4} + x) = \frac{1 + 	an x}{1 - 	an x}$ મળે છે.તેથી,$L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{1 + 	an x}}{{1 - 	an x}}} ight)^{1/x}}$.આ $1^\infty$ સ્વરૂપ છે,તેથી $L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} (\frac{1 + 	an x}{1 - 	an x} - 1)}$.$L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} (\frac{1 + 	an x - 1 + 	an x}{1 - 	an x})} = e^{\mathop {\lim }\limits_{x 	o 0} \frac{2 	an x}{x(1 - 	an x)}}$.કારણ કે $\mathop {\lim }\limits_{x 	o 0} \frac{	an x}{x} = 1$,તેથી $L = e^{2 	imes 1 	imes \frac{1}{1 - 0}} = e^2$.

$\mathop {\lim }\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{1/x}} = $

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