Question

The value of $\lim_{x \to 0} \frac{\cot 4x}{\csc 3x}$ is equal to:

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$
• E. $0$

Hint:

For limits of the form $\frac{\tan mx}{\sin nx}$ or $\frac{\sin mx}{\sin nx}$ as $x \to 0$, the result is always the ratio of the coefficients, $\frac{m}{n}$.

Answer 1

The Correct Option is B

Concept: To evaluate limits involving trigonometric functions as $x \to 0$, it is useful to convert $\cot \theta$ and $\text{cosec } \theta$ into $\sin \theta$ and $\cos \theta$. We then apply the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Step 1: Rewrite the expression using sine and cosine.
\[ \frac{\cot 4x}{\text{cosec } 3x} = \frac{\cos 4x / \sin 4x}{1 / \sin 3x} = \frac{\cos 4x \cdot \sin 3x}{\sin 4x} \]

Step 2: Apply the limit $x \to 0$.
We can separate the terms: \[ \lim_{x \to 0} \left( \cos 4x \right) \cdot \lim_{x \to 0} \left( \frac{\sin 3x}{\sin 4x} \right) \] Since $\cos(0) = 1$, the expression simplifies to: \[ 1 \cdot \lim_{x \to 0} \left( \frac{\sin 3x}{\sin 4x} \right) \]

Step 3: Use standard limit properties to solve.
Multiply and divide by the respective angles $3x$ and $4x$: \[ \lim_{x \to 0} \frac{\frac{\sin 3x}{3x} \cdot 3x}{\frac{\sin 4x}{4x} \cdot 4x} = \frac{\lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3}{\lim_{x \to 0} \frac{\sin 4x}{4x} \cdot 4} \] Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$: \[ \frac{1 \cdot 3}{1 \cdot 4} = \frac{3}{4} \]