If the function f(x) = begin{cases} frac{log_ | vedclass

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(A) For the function to be continuous at $x = 0$,we must have $k = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\ln(1-x+x^2) + \ln(1+x+x^2)}{\sec x - \cos

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(A) For the function to be continuous at $x = 0$,we must have $k = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\ln(1-x+x^2) + \ln(1+x+x^2)}{\sec x - \cos x}$.Using the property $\ln(a) + \ln(b) = \ln(ab)$,the numerator becomes $\ln((1-x+x^2)(1+x+x^2)) = \ln((1+x^2)^2 - x^2) = \ln(1+x^2+x^4)$.The denominator is $\frac{1}{\cos x} - \cos x = \frac{1-\cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$.Thus,$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{\ln(1+x^2+x^4) \cdot \cos x}{\sin^2 x}$.Multiplying and dividing by $(x^2+x^4)$,we get:$\lim_{x 	o 0} \left( \frac{\ln(1+x^2+x^4)}{x^2+x^4} ight) \cdot \left( \frac{x^2+x^4}{\sin^2 x} ight) \cdot \cos x$.Since $\lim_{u 	o 0} \frac{\ln(1+u)}{u} = 1$ and $\lim_{x 	o 0} \frac{\sin x}{x} = 1$,the limit is $1 \cdot 1 \cdot 1 = 1$.Therefore,$k = 1$.

If the function $f(x) = \begin{cases} \frac{\log_{e}(1-x+x^{2}) + \log_{e}(1+x+x^{2})}{\sec x - \cos x}, & x \in (-\frac{\pi}{2}, \frac{\pi}{2}) - \{0\} \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k$ is equal to.

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