limx→0(1sin2x-1sinh2x)=

ltx→01sin2x-1sin h2x =ltx→0(sin h2x-sin2x)x4x4 sinh2x sin2x(00form) =ltx→0sinh2x-sin2xx4 =ltx→0sin h2x-sin 2x4x3+n(n+1)2 =ltx→02 cis h2x-2 cos 2x12x2 =ltx→0sin h2x+sin 2x12x =212+212 =412=13

limx→0(1sin2x-1sinh2x)=

ltx→01sin2x-1sin h2x =ltx→0(sin h2x-sin2x)x4x4 sinh2x sin2x(00form) =ltx→0sinh2x-sin2xx4 =ltx→0sin h2x-sin 2x4x3+n(n+1)2 =ltx→02 cis h2x-2 cos 2x12x2 =ltx→0sin h2x+sin 2x12x =212+212 =412=13

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$\lim_{x \to 0} (\frac{1}{\sin^{2} x} - \frac{1}{\sin h^{2} x}) =$

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$\frac{2}{3}$

$0$

$\frac{1}{3}$

$- \frac{2}{3}$

answer is C.

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Detailed Solution

$\underset{x \to 0}{lt} \frac{1}{\sin^{2} x} - \frac{1}{\sin h^{2} x} = \underset{x \to 0}{lt} \frac{(\sin h^{2} x - \sin^{2} x) x^{4}}{x^{4} \sinh^{2} x \sin^{2} x} (\frac{0}{0} form) = \underset{x \to 0}{lt} \frac{\sinh^{2} x - \sin^{2} x}{x^{4}} = \underset{x \to 0}{lt} \frac{\sin h 2 x - \sin 2 x}{4 x^{3}} + \frac{n (n + 1)}{2} = \underset{x \to 0}{lt} \frac{2 cis h 2 x - 2 \cos 2 x}{12 x^{2}} = \underset{x \to 0}{lt} \frac{\sin h 2 x + \sin 2 x}{12 x} = \frac{2}{12} + \frac{2}{12} = \frac{4}{12} = \frac{1}{3}$