If `Lim_( theta to 0) k theta cosec theta = Lim_( theta to 0) theta cosec k theta , ` prove that k must be ` pm 1 ` :

Answer

To prove that \( k \) must be \( \pm 1 \) given the equation

\[ \lim_{\theta \to 0} k \theta \csc \theta = \lim_{\theta \to 0} \theta \csc(k \theta), \]

we will follow these steps:

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

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

This simplifies to:

\[ \lim_{\theta \to 0} \frac{k \theta}{\sin \theta} = \lim_{\theta \to 0} \frac{\theta}{\sin(k \theta)}. \]

### Step 2: Use the limit property of sine
We know that

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

Thus, we can express our limits in terms of this property:

\[ \lim_{\theta \to 0} \frac{k \theta}{\sin \theta} = k \cdot \lim_{\theta \to 0} \frac{\theta}{\sin \theta} = k \cdot 1 = k, \]

and for the right side:

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

### Step 3: Set the limits equal to each other
Now we have:

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

### Step 4: Solve for \( k \)
Multiplying both sides by \( k \) (assuming \( k \neq 0 \)) gives:

\[ k^2 = 1. \]

Taking the square root of both sides results in:

\[ k = \pm 1. \]

### Conclusion
Thus, we have proven that \( k \) must be \( \pm 1 \).

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