Question

The function $f\left(x\right)=\frac{1+\sin x-\cos x}{1-\sin x-\cos x}$ is not defined at $x=0$. The value of $f\left(0\right)$ so that $f\left(x\right)$ is continuous at $x=0$, is

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

The correct option is B -1

The function will be continuous at $x=0$ if $f\left(0\right)={lim}_{x\to 0}f\left(x\right)$.
So,
$f\left(0\right)=\underset{x\to 0}{lim}\frac{1+\sin x-\cos x}{1-\sin x-\cos x}=\underset{x\to 0}{lim}\frac{(1-\cos x)+\sin x}{(1-\cos x)-\sin x}=\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}=\underset{x\to 0}{lim}\frac{2\sin \frac{x}{2}(\sin \frac{x}{2}+\cos \frac{x}{2})}{2\sin \frac{x}{2}(\sin \frac{x}{2}-\cos \frac{x}{2})}$
Dividing both the numerator and denominator by $2\sin \frac{x}{2}$, since $x\ne 0$, the limit is
$\frac{\sin 0+\cos 0}{\sin 0-\cos 0}=-1$

So, $f\left(0\right)=-1$