Q72 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
$\lim _{\theta \rightarrow \dfrac{\pi}{2}} \dfrac{1-\sin \theta}{\left(\dfrac{\pi}{2}-\theta\right) \cos \theta}$ is equal to
- A$-1$
- B$-\dfrac{1}{2}$
- C$1$
- Correct answerD$\dfrac{1}{2}$
Solution
Let $x = \dfrac{\pi}{2} - \theta$. As $\theta \rightarrow \dfrac{\pi}{2}$, $x \rightarrow 0$.
Substituting $\theta = \dfrac{\pi}{2} - x$, we have $\sin \theta = \sin(\dfrac{\pi}{2} - x) = \cos x$ and $\cos \theta = \cos(\dfrac{\pi}{2} - x) = \sin x$.
The limit becomes $\lim_{x \rightarrow 0} \dfrac{1 - \cos x}{x \sin x}$.
Using the identity $1 - \cos x = 2 \sin^2(\dfrac{x}{2})$, the expression is $\lim_{x \rightarrow 0} \dfrac{2 \sin^2(\dfrac{x}{2})}{x \sin x}$.
Dividing numerator and denominator by $x^2$, we get $\lim_{x \rightarrow 0} \dfrac{2 \left(\dfrac{\sin(x/2)}{x}\right)^2}{\dfrac{\sin x}{x}} = \lim_{x \rightarrow 0} \dfrac{2 \left(\dfrac{1}{2} \cdot \dfrac{\sin(x/2)}{x/2}\right)^2}{\dfrac{\sin x}{x}}$.
Since $\lim_{x \rightarrow 0} \dfrac{\sin x}{x} = 1$, the limit is $\dfrac{2 \cdot (1/2)^2}{1} = \dfrac{2}{4} = \dfrac{1}{2}$.
Answer: $\dfrac{1}{2}$
Substituting $\theta = \dfrac{\pi}{2} - x$, we have $\sin \theta = \sin(\dfrac{\pi}{2} - x) = \cos x$ and $\cos \theta = \cos(\dfrac{\pi}{2} - x) = \sin x$.
The limit becomes $\lim_{x \rightarrow 0} \dfrac{1 - \cos x}{x \sin x}$.
Using the identity $1 - \cos x = 2 \sin^2(\dfrac{x}{2})$, the expression is $\lim_{x \rightarrow 0} \dfrac{2 \sin^2(\dfrac{x}{2})}{x \sin x}$.
Dividing numerator and denominator by $x^2$, we get $\lim_{x \rightarrow 0} \dfrac{2 \left(\dfrac{\sin(x/2)}{x}\right)^2}{\dfrac{\sin x}{x}} = \lim_{x \rightarrow 0} \dfrac{2 \left(\dfrac{1}{2} \cdot \dfrac{\sin(x/2)}{x/2}\right)^2}{\dfrac{\sin x}{x}}$.
Since $\lim_{x \rightarrow 0} \dfrac{\sin x}{x} = 1$, the limit is $\dfrac{2 \cdot (1/2)^2}{1} = \dfrac{2}{4} = \dfrac{1}{2}$.
Answer: $\dfrac{1}{2}$
Hence, the correct answer is option: 'D'.