limx→0 1sin2⁡x−1sin⁡h2x=

limx→0 1sin2⁡x−1sin⁡h2x=limx→0 sin⁡h2x-sin2⁡xsin2⁡xsin⁡h2x =limx→0sinhx+sinxsinhx-sinxsin2xx2sinh2xx2x4 =limx→0sinhx-sinxx3sinhx+sinxx =2limx→0sinhx-sinxx3 =2limx→0coshx-cosx3x2 =2limx→0sinhx+sinx6x=46=23

limx→0 1sin2⁡x−1sin⁡h2x=

limx→0 1sin2⁡x−1sin⁡h2x=limx→0 sin⁡h2x-sin2⁡xsin2⁡xsin⁡h2x =limx→0sinhx+sinxsinhx-sinxsin2xx2sinh2xx2x4 =limx→0sinhx-sinxx3sinhx+sinxx =2limx→0sinhx-sinxx3 =2limx→0coshx-cosx3x2 =2limx→0sinhx+sinx6x=46=23

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

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$\frac{2}{3}$

$\frac{1}{3}$

$\frac{- 2}{3}$

answer is A.

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Detailed Solution

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

$lim_{x \to 0} (\frac{\sin h^{2} x - \sin^{2} x}{\sin^{2} xsin h^{2} x}) = lim_{x \to 0} \frac{(\sin h x + \sin x) (\sin h x - \sin x)}{(\frac{\sin^{2} x}{x^{2}}) (\frac{\sinh^{2} x}{x^{2}}) x^{4}} = lim_{x \to 0} (\frac{\sin h x - \sin x}{x^{3}}) (\frac{\sin h x + \sin x}{x}) = 2 lim_{x \to 0} (\frac{\sin h x - \sin x}{x^{3}}) = 2 lim_{x \to 0} (\frac{\cos h x - \cos x}{3 x^{2}}) = 2 lim_{x \to 0} (\frac{\sin h x + \sin x}{6 x}) = \frac{4}{6} = \frac{2}{3}$