If `Lim_(x to 0) k . cosec x=Lim_(x to 0)x cosec kx`, then k is
(i) `-1,1`
(ii) `-2,2`
(iii) `-(1)/(2),(1)/(2)`
(iv) none of these

Options

A. `-1,1`
B. `-2,2`
C. `-(1)/(2),(1)/(2)`
D. none of these

Answer

To solve the problem, we need to find the value of \( k \) such that:

\[ \lim_{x \to 0} k \cdot \csc x = \lim_{x \to 0} x \cdot \csc(kx) \]

### Step 1: Rewrite the limits using the definition of cosecant
Recall that \( \csc x = \frac{1}{\sin x} \). Therefore, we can rewrite the limits as:

\[ \lim_{x \to 0} k \cdot \frac{1}{\sin x} = \lim_{x \to 0} x \cdot \frac{1}{\sin(kx)} \]

### Step 2: Simplify the limits
This gives us:

\[ \lim_{x \to 0} \frac{k}{\sin x} = \lim_{x \to 0} \frac{x}{\sin(kx)} \]

### Step 3: Use the limit property
We know from calculus that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Thus, we can rewrite the limits as:

\[ \lim_{x \to 0} \frac{k}{\sin x} = k \cdot \lim_{x \to 0} \frac{1}{\sin x/x} = k \cdot 1 = k \]

And for the right side:

\[ \lim_{x \to 0} \frac{x}{\sin(kx)} = \lim_{x \to 0} \frac{1}{\sin(kx)/(kx)} = \frac{1}{k} \]

### Step 4: Set the limits equal to each other
Now we equate both sides:

\[ k = \frac{1}{k} \]

### Step 5: Solve for \( k \)
Multiplying both sides by \( k \) (assuming \( k \neq 0 \)) gives:

\[ k^2 = 1 \]

Taking the square root of both sides, we find:

\[ k = \pm 1 \]

### Conclusion
Thus, the possible values for \( k \) are \( -1 \) and \( 1 \).

### Final Answer
The correct option is (i) \( -1, 1 \).

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