Topic: Engineering Mathematics
$\lim_{\theta\rightarrow\pi/2}\frac{\log(\theta-\pi/2)}{\tan~\theta}$ is
- A. 1
- B. $\pi/2$
- C. $\pi/4$
- D. 0
Answer: D. 0
Explanation: The limit takes the form of $\frac{\infty}{\infty}$. Applying L'Hospital's rule: $\lim_{\theta\rightarrow\pi/2} \frac{\frac{1}{\theta-\pi/2}}{\sec^{2}\theta} = \lim_{\theta\rightarrow\pi/2} \frac{\cos^{2}\theta}{\theta-\pi/2}$. This takes the $\frac{0}{0}$ form. Applying L'Hospital's rule again: $\lim_{\theta\rightarrow\pi/2} \frac{2\cos\theta(-\sin\theta)}{1} = -\sin(2 \times \pi/2) = -\sin\pi = 0$.