Let mathop {Lim}limits_{x to 0} sec^{-1} left | vedclass

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(C) We know that $\mathop {Lim}\limits_{x 	o 0} \frac{\sin x}{x} = 1$,so $\mathop {Lim}\limits_{x 	o 0} \frac{x}{\sin x} = 1$. Thus,$l = \sec^{-1}(1

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$l$ and $m$ both exist

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(C) We know that $\mathop {Lim}\limits_{x 	o 0} \frac{\sin x}{x} = 1$,so $\mathop {Lim}\limits_{x 	o 0} \frac{x}{\sin x} = 1$. Thus,$l = \sec^{-1}(1) = 0$. Similarly,$\mathop {Lim}\limits_{x 	o 0} \frac{	an x}{x} = 1$,so $\mathop {Lim}\limits_{x 	o 0} \frac{x}{	an x} = 1$. Thus,$m = \sec^{-1}(1) = 0$. Since both limits exist and are equal to $0$,$l$ and $m$ both exist.

Let $\mathop {Lim}\limits_{x \to 0} \sec^{-1} \left( \frac{x}{\sin x} \right) = l$ and $\mathop {Lim}\limits_{x \to 0} \sec^{-1} \left( \frac{x}{\tan x} \right) = m$,then

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