Evaluate the following limits :
`lim_(x to pi/2)(e^(sinx)-1)/sinx`

Answer

To evaluate the limit \( \lim_{x \to \frac{\pi}{2}} \frac{e^{\sin x} - 1}{\sin x} \), we can follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{2} \)
First, we substitute \( x \) with \( \frac{\pi}{2} \) in the limit expression.

\[ \sin\left(\frac{\pi}{2}\right) = 1 \]

Thus, we have:

\[ e^{\sin\left(\frac{\pi}{2}\right)} - 1 = e^1 - 1 = e - 1 \]

So, the limit expression becomes:

\[ \frac{e - 1}{1} = e - 1 \]

### Step 2: Evaluate the limit
Now we can write the limit as:

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

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

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

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