If f(x) = frac{1 - sin x + cos x}{1 + sin x + | vedclass

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(A) Since $f$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.Using $L$'$H$ôpital's rule for the limit $\lim_{x \rightarro

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$-1$

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(A) Since $f$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.Using $L$'$H$ôpital's rule for the limit $\lim_{x \rightarrow \pi} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$:$\lim_{x \rightarrow \pi} \frac{\frac{d}{dx}(1 - \sin x + \cos x)}{\frac{d}{dx}(1 + \sin x + \cos x)} = \lim_{x \rightarrow \pi} \frac{-\cos x - \sin x}{\cos x - \sin x}$.Substituting $x = \pi$:$f(\pi) = \frac{-\cos(\pi) - \sin(\pi)}{\cos(\pi) - \sin(\pi)} = \frac{-(-1) - 0}{-1 - 0} = \frac{1}{-1} = -1$.

If $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ for $x \neq \pi$ is continuous at $x = \pi$,then $f(\pi) =$

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