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

Answer

To evaluate the limit

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

we will follow these steps:

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

\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad e^{\cos\left(\frac{\pi}{2}\right)} = e^0 = 1. \]

Thus, we have:

\[ \frac{e^{\cos\left(\frac{\pi}{2}\right)} - 1}{\cos\left(\frac{\pi}{2}\right)} = \frac{1 - 1}{0} = \frac{0}{0}. \]

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and denominator with respect to \( x \):

- The derivative of the numerator \( e^{\cos x} - 1 \) is \( -\sin x \cdot e^{\cos x} \) (using the chain rule).
- The derivative of the denominator \( \cos x \) is \( -\sin x \).

Thus, we rewrite the limit as:

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

### Step 3: Evaluate the new limit
Now we can substitute \( x = \frac{\pi}{2} \) again into the simplified expression:

\[ \cos\left(\frac{\pi}{2}\right) = 0 \implies e^{\cos\left(\frac{\pi}{2}\right)} = e^0 = 1. \]

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

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