Evaluate the following limits :
`lim_(x to -1)(x^(3)+1)/(x+1)`.

Answer

To evaluate the limit \(\lim_{x \to -1} \frac{x^3 + 1}{x + 1}\), we can follow these steps:

### Step 1: Identify the Indeterminate Form
First, we substitute \(x = -1\) into the expression:
\[ \frac{(-1)^3 + 1}{-1 + 1} = \frac{-1 + 1}{0} = \frac{0}{0} \]
This is an indeterminate form, so we need to simplify the expression.

### Step 2: Factor the Numerator
We can use the sum of cubes formula, which states that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
In our case, \(a = x\) and \(b = 1\). Thus, we can rewrite \(x^3 + 1\) as:
\[ x^3 + 1 = (x + 1)(x^2 - x + 1) \]

### Step 3: Rewrite the Limit
Now we can substitute this factorization back into our limit:
\[ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} = \lim_{x \to -1} \frac{(x + 1)(x^2 - x + 1)}{x + 1} \]

### Step 4: Cancel the Common Factors
Since \(x + 1\) is a common factor in the numerator and denominator, we can cancel it (as long as \(x \neq -1\)):
\[ \lim_{x \to -1} (x^2 - x + 1) \]

### Step 5: Evaluate the Limit
Now we can directly substitute \(x = -1\):
\[ (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} = 3 \]