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

Answer

To evaluate the limit

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

we will follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{2} \)

First, we substitute \( x = \frac{\pi}{2} \) into the limit expression.

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

So we have:

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

### Step 2: Evaluate the denominator

Now, we evaluate the denominator:

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

### Step 3: Substitute into the limit expression

Substituting these values into the limit expression gives:

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

### Conclusion

Thus, the limit evaluates to:

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