Evaluate the limits, if exist `lim_(x rarr 0)(x(e^x-1))/(1-cosx)`.

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{x(e^x - 1)}{1 - \cos x}, \]

we will follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), both the numerator and denominator approach 0. Therefore, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Rewrite the limit
We can rewrite the limit as:

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

### Step 3: Simplify the limit
We can separate this into two limits:

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

### Step 4: Evaluate the first limit
The first limit can be evaluated using the known limit:

\[ \lim_{x \to 0} \frac{e^x - 1}{x} = 1. \]

### Step 5: Evaluate the second limit
For the second limit, we can use the half-angle formula for cosine:

\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right). \]

Thus, we rewrite the limit:

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

### Step 6: Rewrite using sine limit
Now, we can express this as:

\[ \frac{1}{2} \lim_{x \to 0} \frac{x^2}{\sin^2\left(\frac{x}{2}\right)}. \]

Using the fact that

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

we have:

\[ \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} = 1 \implies \lim_{x \to 0} \frac{x^2}{\sin^2\left(\frac{x}{2}\right)} = \lim_{x \to 0} \left(\frac{x}{\frac{x}{2}}\right)^2 = 4. \]

### Step 7: Combine the results
Now we can combine the results of both limits:

\[ \lim_{x \to 0} \frac{x(e^x - 1)}{1 - \cos x} = 1 \cdot \frac{1}{2} \cdot 4 = 2. \]

### Final Answer
Thus, the limit is

\[ \boxed{2}. \]