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 can break it down into manageable steps.

### Step 1: Rewrite the Limit
We start by rewriting the limit expression:

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

### Step 2: Split the Limit
We can split the limit into two parts:

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

### Step 3: Evaluate Each Limit
Now we evaluate each limit separately.

1. **For the first limit**:

We know that

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

Therefore,

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

2. **For the second limit**:

We also know that

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

### Step 4: Combine the Results
Now we combine the results from the two limits:

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

### Final Result
Thus, the limit is

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

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