mathop {lim }limits_{x to 0} {(cos ax)^{csc^2 | vedclass

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Question

(A) આપેલ લક્ષ $1^\infty$ સ્વરૂપનું છે. સૂત્ર $\mathop {\lim }\limits_{x 	o 0} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o 0} (f(x) - 1)g(x)}$ નો

Vedclass · Accepted Answer

$e^{\left( {\frac{{ - {a^2}}}{{2{b^2}}}} \right)}$

Vedclass · Answer

(A) આપેલ લક્ષ $1^\infty$ સ્વરૂપનું છે. સૂત્ર $\mathop {\lim }\limits_{x 	o 0} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o 0} (f(x) - 1)g(x)}$ નો ઉપયોગ કરતા: $L = e^{\mathop {\lim }\limits_{x 	o 0} (\cos ax - 1) \csc^2 bx}$ $L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{-(1 - \cos ax)}{\sin^2 bx}}$ $L = e^{\mathop {\lim }\limits_{x 	o 0} -\left( \frac{1 - \cos ax}{(ax)^2} ight) \cdot \frac{(ax)^2}{\left( \frac{\sin bx}{bx} \cdot bx ight)^2}}$ કારણ કે $\mathop {\lim }\limits_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$ અને $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$: $L = e^{-\frac{1}{2} \cdot \frac{a^2}{b^2}} = e^{\left( -\frac{a^2}{2b^2} ight)}$

$\mathop {\lim }\limits_{x \to 0} {(\cos ax)^{\csc^2 bx}}$ ની કિંમત શોધો.

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