`lim_(xrarr0)("cosec"x-cotx)/(x)` is equal to

Options

A. `1/2`
B. 1
C. `1/2`
D. 1

Explanation

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

### Step 1: Rewrite the functions
We start by rewriting \(\csc x\) and \(\cot x\) in terms of sine and cosine:
\[ \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x} \]
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: Identify the indeterminate form
As \(x\) approaches \(0\), both the numerator \(1 - \cos x\) and the denominator \(x \sin x\) approach \(0\). This gives us the indeterminate form \(\frac{0}{0}\).

### Step 4: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit on the right exists.

#### Derivative of the numerator:
The derivative of \(1 - \cos x\) is \(\sin x\).

#### Derivative of the denominator:
The derivative of \(x \sin x\) can be found using the product rule:
\[ \frac{d}{dx}(x \sin x) = \sin x + x \cos x \]

### Step 5: Rewrite the limit using derivatives
Now we can rewrite our limit:
\[ \lim_{x \to 0} \frac{\sin x}{\sin x + x \cos x} \]

### Step 6: Substitute \(x = 0\)
Now we substitute \(x = 0\):
- The numerator \(\sin(0) = 0\)
- The denominator \(\sin(0) + 0 \cdot \cos(0) = 0 + 0 = 0\)

This again gives us the indeterminate form \(\frac{0}{0}\), so we apply L'Hôpital's Rule again.

### Step 7: Apply L'Hôpital's Rule again
#### Derivative of the numerator:
The derivative of \(\sin x\) is \(\cos x\).

#### Derivative of the denominator:
The derivative of \(\sin x + x \cos x\) is:
\[ \cos x + (\cos x - x \sin x) = 2\cos x - x \sin x \]

### Step 8: Rewrite the limit again
Now we have:
\[ \lim_{x \to 0} \frac{\cos x}{2\cos x - x \sin x} \]

### Step 9: Substitute \(x = 0\) again
Substituting \(x = 0\):
- The numerator \(\cos(0) = 1\)
- The denominator \(2\cos(0) - 0 \cdot \sin(0) = 2 - 0 = 2\)

Thus, we have:
\[ \lim_{x \to 0} \frac{\cos x}{2\cos x - x \sin x} = \frac{1}{2} \]

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

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