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

Answer

To evaluate the limit

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

we can follow these steps:

### Step 1: Recognize the form of the limit
As \( x \) approaches 0, both the numerator \( e^{\sin x} - 1 \) and the denominator \( x \) approach 0. This gives us a \( \frac{0}{0} \) indeterminate form, which allows us to apply L'Hôpital's Rule or manipulate the expression.

**Hint:** Check if the limit results in an indeterminate form to decide the next steps.

### Step 2: Rewrite the expression
We can rewrite the limit by multiplying and dividing by \( \sin x \):

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

**Hint:** This step helps to separate the limit into two parts, making it easier to evaluate.

### Step 3: Evaluate the first limit
Now we evaluate the first limit:

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

We can recognize that as \( x \to 0 \), \( \sin x \to 0 \) as well. This limit is of the form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule:

1. Differentiate the numerator: the derivative of \( e^{\sin x} \) is \( e^{\sin x} \cos x \).
2. Differentiate the denominator: the derivative of \( \sin x \) is \( \cos x \).

Thus, we have:

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

**Hint:** Use L'Hôpital's Rule when you encounter \( \frac{0}{0} \) forms.

### Step 4: Evaluate the second limit
Now we evaluate the second limit:

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

This is a well-known limit, and it equals 1.

**Hint:** Remember that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

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

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

### Final Answer
Thus, the limit is

\[ \boxed{1}. \]