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

Answer

To evaluate the limit

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

we will follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the limit expression:

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

This gives us an indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter \( \frac{0}{0} \), it indicates that further analysis is needed, such as applying L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator:

\[ \lim_{x \to 0} \frac{e^{\sin x} - 1}{\sin x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^{\sin x})}{\frac{d}{dx}(\sin x)}. \]

### Step 3: Differentiate the Numerator and Denominator

Now we differentiate the numerator and the denominator:

- The derivative of the numerator \( e^{\sin x} \) using the chain rule is:

\[ \frac{d}{dx}(e^{\sin x}) = e^{\sin x} \cdot \cos x. \]

- The derivative of the denominator \( \sin x \) is:

\[ \frac{d}{dx}(\sin x) = \cos x. \]

### Step 4: Rewrite the Limit

Now we can rewrite the limit using the derivatives:

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

### Step 5: Simplify the Expression

We can simplify the expression since \( \cos x \) in the numerator and denominator cancels out (as long as \( \cos x \neq 0 \), which is true near \( x = 0 \)):

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

### Step 6: Substitute \( x = 0 \) Again

Now we substitute \( x = 0 \):

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

### Conclusion

Thus, the limit evaluates to:

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

**Final Answer:** \( 1 \)

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