Understanding limits in rational functions is crucial for calculus. When evaluating limits, if the denominator does not equal zero at a specific point, you can directly substitute that value into the function. For example, consider the function:

$$\frac{x^2 + 3x + 2}{x + 1}$$

To find the limit as \(x\) approaches 0, simply substitute 0 into the function:

$$\frac{0^2 + 3(0) + 2}{0 + 1} = \frac{2}{1} = 2$$

However, if you want to find the limit as \(x\) approaches -1, you first check the denominator:

$$x + 1 = -1 + 1 = 0$$

Since the denominator equals zero, you cannot substitute directly. Instead, you should factor both the numerator and the denominator. The numerator \(x^2 + 3x + 2\) factors to:

$$ (x + 1)(x + 2) $$

Thus, the function can be rewritten as:

$$\frac{(x + 1)(x + 2)}{(x + 1)}$$

Now, you can cancel the common factor \(x + 1\), leading to:

$$x + 2$$

Now, you can evaluate the limit as \(x\) approaches -1:

$$-1 + 2 = 1$$

In summary, when faced with a rational function where the denominator equals zero, the steps are:

1. Check if the denominator equals zero at the limit point.
2. If it does, factor both the numerator and denominator.
3. Cancel any common factors.
4. Evaluate the limit using the simplified function.

For another example, consider:

$$\frac{x^2 + 2x - 15}{x - 3}$$

To find the limit as \(x\) approaches 3, check the denominator:

$$3 - 3 = 0$$

Next, factor the numerator, which factors to:

$$ (x - 3)(x + 5) $$

Now, the function becomes:

$$\frac{(x - 3)(x + 5)}{(x - 3)}$$

Cancel the common factor \(x - 3\) to get:

$$x + 5$$

Now, evaluate the limit as \(x\) approaches 3:

$$3 + 5 = 8$$

Lastly, consider:

$$\frac{x + 2}{x^2 - x - 6}$$

To find the limit as \(x\) approaches -2, check the denominator:

$$(-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0$$

Factor the denominator \(x^2 - x - 6\) to:

$$ (x + 2)(x - 3) $$

Now the function is:

$$\frac{x + 2}{(x + 2)(x - 3)}$$

Cancel the common factor \(x + 2\) to simplify to:

$$\frac{1}{x - 3}$$

Now, evaluate the limit as \(x\) approaches -2:

$$\frac{1}{-2 - 3} = \frac{1}{-5} = -\frac{1}{5}$$

By following these steps, you can effectively find limits for rational functions, even when the denominator approaches zero.