Evaluate the following limits :
`Lim_(x to 0 ) (" cosec " x - cot x)/x `

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{\csc x - \cot x}{x} \), we can follow these steps:

### Step 1: Rewrite the functions
Recall that:
\[ \csc x = \frac{1}{\sin x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \]
Thus, we can rewrite the expression:
\[ \csc x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x} \]

### Step 2: Substitute into the limit
Now, substitute this back into the limit:
\[ \lim_{x \to 0} \frac{\csc x - \cot x}{x} = \lim_{x \to 0} \frac{\frac{1 - \cos x}{\sin x}}{x} = \lim_{x \to 0} \frac{1 - \cos x}{x \sin x} \]

### Step 3: Use the identity for \(1 - \cos x\)
We know that:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
Substituting this into our limit gives:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x \sin x} \]

### Step 4: Rewrite \( \sin x \)
Using the double angle identity, we can express \( \sin x \) as:
\[ \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]
Now substitute this into the limit:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x \cdot 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)} = \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{x \cos\left(\frac{x}{2}\right)} \]

### Step 5: Simplify the limit
We can express \( x \) as \( 2 \cdot \frac{x}{2} \):
\[ = \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2} \cdot 2 \cos\left(\frac{x}{2}\right)} = \frac{1}{2} \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \frac{1}{\cos\left(\frac{x}{2}\right)} \]

### Step 6: Evaluate the limit
As \( x \to 0 \):
\[ \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} = 1 \quad \text{and} \quad \lim_{x \to 0} \cos\left(\frac{x}{2}\right) = 1 \]
Thus, we have:
\[ \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{\csc x - \cot x}{x} = \frac{1}{2} \]