Compute `lim_(x rarr 0)(e^(3x)-sinx-1)/x`

Explanation

To compute the limit

\[ \lim_{x \to 0} \frac{e^{3x} - \sin x - 1}{x}, \]

we will follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \(x = 0\) into the expression:

\[ e^{3 \cdot 0} - \sin(0) - 1 = 1 - 0 - 1 = 0. \]

Thus, the limit takes the form \(\frac{0}{0}\), which is an indeterminate form.

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

\[ \lim_{x \to 0} \left( \frac{e^{3x} - 1}{x} - \frac{\sin x}{x} \right). \]

### Step 3: Separate the limits
Using the properties of limits, we can separate this into two limits:

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

### Step 4: Evaluate the first limit
For the first limit, we can use the standard result:

\[ \lim_{x \to 0} \frac{e^{kx} - 1}{x} = k, \]

where \(k = 3\) in our case. Thus,

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

### Step 5: Evaluate the second limit
For the second limit, we use the standard result:

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

### Step 6: Combine the results
Now we combine the results from the two limits:

\[ 3 - 1 = 2. \]

### Final Answer
Thus, we have:

\[ \lim_{x \to 0} \frac{e^{3x} - \sin x - 1}{x} = 2. \]