If a 0 and b

Question

(A) Given the limit: $L = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$Using the identity $1 - \cos 2	heta = 2\si

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$\frac{a\sqrt{2}}{b}$

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(A) Given the limit: $L = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$Using the identity $1 - \cos 2	heta = 2\sin^2 	heta$,we have:$L = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt {2\sin^2 ax} }}{{\sin bx}} = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt{2} |\sin ax|}}{{\sin bx}}$Since $x 	o 0^+$,$ax > 0$ (as $a > 0$),so $|\sin ax| = \sin ax$.$L = \sqrt{2} \mathop {\lim }\limits_{x 	o {0^ + }} \frac{\sin ax}{\sin bx} = \sqrt{2} \mathop {\lim }\limits_{x 	o {0^ + }} \left( \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{ax}{bx} ight)$$L = \sqrt{2} \cdot 1 \cdot 1 \cdot \frac{a}{b} = \frac{a\sqrt{2}}{b}$

If $a > 0$ and $b < 0$,then $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$ is equal to:

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