Q108 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\lim _\limits{x \rightarrow 0}\left(\dfrac{\sin a x}{\sin b x}\right)^k \text { equals }$
- A$\dfrac{b}{a}$
- B$\left(\dfrac{b}{a}\right)^k$
- Correct answerC$\left(\dfrac{a}{b}\right)^k$
- D$\dfrac{a}{b}$
Solution
The given limit is $L = \lim_{x \rightarrow 0} \left( \dfrac{\sin ax}{\sin bx} \right)^k$.
We can rewrite the expression inside the limit as:
$\dfrac{\sin ax}{\sin bx} = \dfrac{\sin ax}{ax} \cdot \dfrac{bx}{\sin bx} \cdot \dfrac{ax}{bx}$
Using the standard limit $\lim_{\theta \rightarrow 0} \dfrac{\sin \theta}{\theta} = 1$, we have:
$\lim_{x \rightarrow 0} \dfrac{\sin ax}{ax} = 1$ and $\lim_{x \rightarrow 0} \dfrac{bx}{\sin bx} = 1$.
Therefore, $\lim_{x \rightarrow 0} \dfrac{\sin ax}{\sin bx} = \lim_{x \rightarrow 0} \left( \dfrac{\sin ax}{ax} \cdot \dfrac{bx}{\sin bx} \cdot \dfrac{a}{b} \right) = 1 \cdot 1 \cdot \dfrac{a}{b} = \dfrac{a}{b}$.
Applying the power $k$ to the limit:
$L = \left( \lim_{x \rightarrow 0} \dfrac{\sin ax}{\sin bx} \right)^k = \left( \dfrac{a}{b} \right)^k$.
Answer: $\left(\dfrac{a}{b}\right)^k$
We can rewrite the expression inside the limit as:
$\dfrac{\sin ax}{\sin bx} = \dfrac{\sin ax}{ax} \cdot \dfrac{bx}{\sin bx} \cdot \dfrac{ax}{bx}$
Using the standard limit $\lim_{\theta \rightarrow 0} \dfrac{\sin \theta}{\theta} = 1$, we have:
$\lim_{x \rightarrow 0} \dfrac{\sin ax}{ax} = 1$ and $\lim_{x \rightarrow 0} \dfrac{bx}{\sin bx} = 1$.
Therefore, $\lim_{x \rightarrow 0} \dfrac{\sin ax}{\sin bx} = \lim_{x \rightarrow 0} \left( \dfrac{\sin ax}{ax} \cdot \dfrac{bx}{\sin bx} \cdot \dfrac{a}{b} \right) = 1 \cdot 1 \cdot \dfrac{a}{b} = \dfrac{a}{b}$.
Applying the power $k$ to the limit:
$L = \left( \lim_{x \rightarrow 0} \dfrac{\sin ax}{\sin bx} \right)^k = \left( \dfrac{a}{b} \right)^k$.
Answer: $\left(\dfrac{a}{b}\right)^k$
Hence, the correct answer is option: 'C'.