Evaluate `lim_(x to 0) (x^(3) cot x)/(1 - cos x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}, \]

we will follow a systematic approach using known limits.

### Step 1: Rewrite the expression

We know that

\[ \cot x = \frac{\cos x}{\sin x}. \]

So we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{x^3 \cdot \frac{\cos x}{\sin x}}{1 - \cos x} = \lim_{x \to 0} \frac{x^3 \cos x}{\sin x (1 - \cos x)}. \]

### Step 2: Analyze the limit

As \( x \to 0 \), both the numerator and denominator approach 0, which gives us an indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule or manipulate the expression further.

### Step 3: Use known limits

We will use the known limits:

1. \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)
2. \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\)

We can rewrite the limit using these known properties:

\[ \lim_{x \to 0} \frac{x^3 \cos x}{\sin x (1 - \cos x)} = \lim_{x \to 0} \frac{x^3}{\sin x} \cdot \frac{\cos x}{1 - \cos x}. \]

### Step 4: Simplify the limit

Now we can separate the limit:

\[ = \lim_{x \to 0} \frac{x^3}{\sin x} \cdot \lim_{x \to 0} \frac{\cos x}{1 - \cos x}. \]

### Step 5: Evaluate each limit

1. For \(\lim_{x \to 0} \frac{x^3}{\sin x}\):

Using \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), we have:

\[ \lim_{x \to 0} \frac{x^3}{\sin x} = \lim_{x \to 0} x^3 \cdot \frac{1}{\sin x/x} = 0. \]

2. For \(\lim_{x \to 0} \frac{\cos x}{1 - \cos x}\):

Using the limit \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\), we rewrite it as:

\[ \lim_{x \to 0} \frac{\cos x}{1 - \cos x} = \lim_{x \to 0} \frac{\cos x}{\frac{1 - \cos x}{x^2} \cdot x^2} = \lim_{x \to 0} \frac{\cos x \cdot x^2}{\frac{1}{2}} = 0. \]

### Step 6: Combine the results

Now, we can combine the results:

\[ \lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x} = 0 \cdot \text{(some finite value)} = 0. \]

Thus, the final answer is:

\[ \lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x} = 0. \]