Evaluate the following limits :
`Lim_(x to 0) (e^(ax)-1)/(sin x)`

Answer

To evaluate the limit

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

we can use some standard limits. Let's go through the steps systematically.

### Step 1: Identify Standard Limits
We will use the following standard limits:
1. \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)
2. \(\lim_{x \to 0} \frac{e^{ax} - 1}{ax} = 1\)

### Step 2: Rewrite the Expression
To use these limits, we need to manipulate the expression. We can rewrite the limit as follows:

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

### Step 3: Separate the Limits
Now we can separate the limit into two parts:

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

### Step 4: Evaluate Each Limit
1. For the first limit, using the standard limit we know:

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

2. For the second limit, we can rewrite it using the standard limit:

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

### Step 5: Combine the Results
Now we can combine the results from both limits:

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

### Final Answer
Thus, the limit evaluates to:

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

---