Question
The function $$f\left( x \right) = \frac{{1 - \sin \,x + \cos \,x}}{{1 + \sin \,x + \cos \,x}}$$ is not defined at $$x = \pi .$$ The value of $$f\left( \pi \right)$$ so that $$f\left( x \right)$$ is continuous at $$x = \pi ,$$ is :
A.
$$ - \frac{1}{2}$$
B.
$$\frac{1}{2}$$
C.
$$ - 1$$
D.
$$1$$
Answer :
$$ - 1$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to \pi } \frac{{1 - \sin \,x + \cos \,x}}{{1 + \sin \,x + \cos \,x}} \cr
& {\text{Using L'hospital's rules }} \cr
& \Rightarrow \mathop {\lim }\limits_{x \to \pi } \frac{{ - \cos \,x - \sin \,x}}{{\cos \,x - \sin \,x}} = \frac{{ - \cos \,\pi - \sin \,\pi }}{{\cos \,\pi - \sin \,\pi }} = \frac{{ - \left( { - 1} \right) - 0}}{{ - 1 - 0}} = - 1 \cr} $$