Question

If [.] denotes the greatest integer function, then $\underset{x\to 0}{lim}\frac{\sin [-{\sec }^{2}x]}{1+\left[\cos x\right]}$ is?

• A. $\sin 1$
• B. $-\sin 1$
• C. $\sin 2$
• D. $-\sin 2$

Answer 1

Answer: (D)

$-\sin 2$

$\underset{x\to 0}{lim}\frac{\sin [-{\sec }^{2}x]}{1+\left[\cos x\right]}=\frac{\sin [-(1^{+}{)}^2]}{1+0}=\sin (-2)=-\sin 2$
Since $-(1^{+}{)}^2\in (-2,-1)\Rightarrow [-\left(1^{+}{)}^2\right]=-2$