Evaluate the following Limits
`lim_(xto pi)(1+sec^(3)x)/(tan^(2)x)`

Answer

To evaluate the limit

\[ \lim_{x \to \pi} \frac{1 + \sec^3 x}{\tan^2 x} \]

we will follow these steps:

### Step 1: Direct Substitution
First, we substitute \( x = \pi \) into the expression.

\[ \tan(\pi) = 0 \quad \text{and} \quad \sec(\pi) = -1 \]

Thus,

\[ 1 + \sec^3(\pi) = 1 + (-1)^3 = 1 - 1 = 0 \]

This gives us:

\[ \frac{1 + \sec^3(\pi)}{\tan^2(\pi)} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Applying L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

**Numerator:**
\[ \frac{d}{dx}(1 + \sec^3 x) = 0 + 3\sec^2 x \cdot \sec x \tan x = 3\sec^3 x \tan x \]

**Denominator:**
\[ \frac{d}{dx}(\tan^2 x) = 2\tan x \cdot \sec^2 x \]

Now, we rewrite the limit using these derivatives:

\[ \lim_{x \to \pi} \frac{3\sec^3 x \tan x}{2\tan x \sec^2 x} \]

### Step 3: Simplifying the Expression
We can cancel \(\tan x\) from the numerator and denominator (as long as \(x \neq \pi\)):

\[ \lim_{x \to \pi} \frac{3\sec^3 x}{2\sec^2 x} = \lim_{x \to \pi} \frac{3\sec x}{2} \]

### Step 4: Evaluating the Limit
Now we substitute \(x = \pi\):

\[ \sec(\pi) = -1 \]

Thus,

\[ \lim_{x \to \pi} \frac{3\sec x}{2} = \frac{3(-1)}{2} = -\frac{3}{2} \]

### Final Answer
The limit is

\[ \boxed{-\frac{3}{2}} \]