Discuss the continuity of the function
`f(x) ={:{((1+cos x)/(tan^2 x) ", "x ne pi),((1)/(2) ", " x=pi):},`
`at x=pi`.

Answer

To discuss the continuity of the function

\[ f(x) = \begin{cases} \frac{1 + \cos x}{\tan^2 x} & \text{if } x \neq \pi \\ \frac{1}{2} & \text{if } x = \pi \end{cases} \]

at \( x = \pi \), we need to check the following conditions:

1. **Existence of the limit** as \( x \) approaches \( \pi \).
2. **Value of the function** at \( x = \pi \).
3. **Equality of the limit and the function value** at \( x = \pi \).

### Step 1: Compute the limit as \( x \) approaches \( \pi \)

We need to find:

\[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x} \]

Substituting \( x = \pi \):

- \( \cos(\pi) = -1 \)
- \( \tan(\pi) = 0 \)

Thus, we have:

\[ 1 + \cos(\pi) = 1 - 1 = 0 \]
\[ \tan^2(\pi) = 0^2 = 0 \]

This gives us the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator:

- The derivative of the numerator \( 1 + \cos x \) is \( -\sin x \).
- The derivative of the denominator \( \tan^2 x \) is \( 2\tan x \cdot \sec^2 x \).

Now we can rewrite the limit:

\[ \lim_{x \to \pi} \frac{-\sin x}{2\tan x \sec^2 x} \]

### Step 3: Substitute \( x = \pi \) again

Substituting \( x = \pi \):

- \( \sin(\pi) = 0 \)
- \( \tan(\pi) = 0 \)
- \( \sec^2(\pi) = \frac{1}{\cos^2(\pi)} = 1 \)

This again gives us \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again

Differentiate the numerator and denominator again:

- The derivative of \( -\sin x \) is \( -\cos x \).
- The derivative of \( 2\tan x \sec^2 x \) can be computed using the product rule.

After differentiating, we can simplify and evaluate the limit again.

### Step 5: Evaluate the limit

Continuing from the previous step, after simplification, we find:

\[ \lim_{x \to \pi} \frac{-\cos x}{2(\sec^2 x + 2\tan^2 x \sec^2 x)} \]

Substituting \( x = \pi \):

- \( -\cos(\pi) = 1 \)
- The denominator becomes a non-zero value.

Thus, we find:

\[ \lim_{x \to \pi} f(x) = \frac{1}{2} \]

### Step 6: Check the value of the function at \( x = \pi \)

From the definition of the function:

\[ f(\pi) = \frac{1}{2} \]

### Step 7: Conclusion

Since:

\[ \lim_{x \to \pi} f(x) = f(\pi) = \frac{1}{2} \]

We conclude that \( f(x) \) is continuous at \( x = \pi \).