The value of limx→0log1+x+x2+log1-x+x2sec x-cos x is

limx→0log1+x22-x2sin xtan x =limx→0log1+x2+x4sin xtan x =limx→0log1+x2+x4sin xxtan xxx2 =limx→0log1+x2+x4x2 =limx→02x+4x31+x2+x42x =limx→01+2x21+x2+x4=1

The value of limx→0log1+x+x2+log1-x+x2sec x-cos x is

limx→0log1+x22-x2sin xtan x =limx→0log1+x2+x4sin xtan x =limx→0log1+x2+x4sin xxtan xxx2 =limx→0log1+x2+x4x2 =limx→02x+4x31+x2+x42x =limx→01+2x21+x2+x4=1

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The value of $\lim_{x \to 0} \frac{\log (1 + x + x^{2}) + \log (1 - x + x^{2})}{\sec x - \cos x}$ is

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answer is B.

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Detailed Solution

$\lim_{x \to 0} \frac{\log [{(1 + x^{2})}^{2} - x^{2}]}{\sin xtan x} = \lim_{x \to 0} \frac{\log (1 + x^{2} + x^{4})}{\sin xtan x} = \lim_{x \to 0} \frac{\log (1 + x^{2} + x^{4})}{\frac{\sin x}{x} \frac{\tan x}{x} x^{2}} = \lim_{x \to 0} \frac{\log (1 + x^{2} + x^{4})}{x^{2}} = \lim_{x \to 0} \frac{\frac{2 x + 4 x^{3}}{1 + x^{2} + x^{4}}}{2 x} = \lim_{x \to 0} \frac{1 + 2 x^{2}}{1 + x^{2} + x^{4}} = 1$