The value of mathop {lim }limits_{x to 0} {(c | vedclass

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Question

(A) The given limit is of the form $1^\infty$. Using the formula $\mathop {\lim }\limits_{x 	o 0} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o 0}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given limit is of the form $1^\infty$. Using the formula $\mathop {\lim }\limits_{x 	o 0} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o 0} (f(x) - 1)g(x)}$,we have: $L = e^{\mathop {\lim }\limits_{x 	o 0} (\cos ax - 1) \csc^2 bx}$ $L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{-(1 - \cos ax)}{\sin^2 bx}}$ $L = e^{\mathop {\lim }\limits_{x 	o 0} -\left( \frac{1 - \cos ax}{(ax)^2} ight) \cdot \frac{(ax)^2}{\left( \frac{\sin bx}{bx} \cdot bx ight)^2}}$ Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$: $L = e^{-\frac{1}{2} \cdot \frac{a^2}{b^2}} = e^{\left( -\frac{a^2}{2b^2} ight)}$

The value of $\mathop {\lim }\limits_{x \to 0} {(\cos ax)^{\csc^2 bx}}$ is-

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