If limx→0 xn−sin⁡xnx−sinn⁡x is non –zero finite, then n must be equal to

For n = 0, we have limx→0 1−sin⁡1x−1=sin⁡1−1 For n=1,limx→0 x−sin⁡xx−sin⁡x=1 For n=2,limx→0 x2−sin2⁡xx−sin2⁡x =limx→0 1−sin2⁡xx21x−sin2⁡xx2This does not exist for n=3 also givenlimit does not exist. Hence, n=0 or 1

If limx→0 xn−sin⁡xnx−sinn⁡x is non –zero finite, then n must be equal to

For n = 0, we have limx→0 1−sin⁡1x−1=sin⁡1−1 For n=1,limx→0 x−sin⁡xx−sin⁡x=1 For n=2,limx→0 x2−sin2⁡xx−sin2⁡x =limx→0 1−sin2⁡xx21x−sin2⁡xx2This does not exist for n=3 also givenlimit does not exist. Hence, n=0 or 1

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If $lim_{x \to 0} \frac{x^{n} - \sin x^{n}}{x - \sin^{n} x}$ is non –zero finite, then n must be equal to

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answer is B.

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Detailed Solution

For n = 0, we have $lim_{x \to 0} \frac{1 - \sin 1}{x - 1} = \sin 1 - 1$

$For n = 1, lim_{x \to 0} \frac{x - \sin x}{x - \sin x} = 1$

$For n = 2, lim_{x \to 0} \frac{x^{2} - \sin^{2} x}{x - \sin^{2} x} = lim_{x \to 0} \frac{1 - \frac{\sin^{2} x}{x^{2}}}{\frac{1}{x} - \frac{\sin^{2} x}{x^{2}}}$