Q93 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { The value of } \lim _\limits{x \rightarrow 0} \dfrac{\sin (a+x)-\sin (a-x)}{x} \text { is }$
- A1
- Correct answerB$2 \cos a$
- C0
- D$2 \sin a$
Solution
The given limit is $L = \lim_{x \to 0} \dfrac{\sin(a+x) - \sin(a-x)}{x}$.
Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \dfrac{C+D}{2} \right) \sin \left( \dfrac{C-D}{2} \right)$, we set $C = a+x$ and $D = a-x$.
Then $\dfrac{C+D}{2} = \dfrac{a+x+a-x}{2} = a$ and $\dfrac{C-D}{2} = \dfrac{a+x-(a-x)}{2} = x$.
Substituting these into the expression, we get $L = \lim_{x \to 0} \dfrac{2 \cos a \sin x}{x}$.
Since $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$, the expression simplifies to $2 \cos a \times 1 = 2 \cos a$.
Answer: $2 \cos a$
Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \dfrac{C+D}{2} \right) \sin \left( \dfrac{C-D}{2} \right)$, we set $C = a+x$ and $D = a-x$.
Then $\dfrac{C+D}{2} = \dfrac{a+x+a-x}{2} = a$ and $\dfrac{C-D}{2} = \dfrac{a+x-(a-x)}{2} = x$.
Substituting these into the expression, we get $L = \lim_{x \to 0} \dfrac{2 \cos a \sin x}{x}$.
Since $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$, the expression simplifies to $2 \cos a \times 1 = 2 \cos a$.
Answer: $2 \cos a$
Hence, the correct answer is option: 'B'.