Understanding how to find the limit of a rational function is essential in calculus, especially when dealing with cases where the denominator approaches zero. Let's explore the process through two examples that illustrate different scenarios.

In the first example, we need to find the limit of the function \( \frac{x^3 - 2x^2 + x}{x - 1} \) as \( x \) approaches 1. Substituting \( x = 1 \) directly into the denominator results in \( 1 - 1 = 0 \), indicating that we must factor the function to resolve the indeterminate form.

To factor the numerator, we first notice that each term contains an \( x \), allowing us to factor out \( x \):

\( x(x^2 - 2x + 1) \).

Next, we recognize that \( x^2 - 2x + 1 \) is a perfect square, which can be factored further as \( (x - 1)^2 \). Thus, the entire expression becomes:

\( \frac{x(x - 1)^2}{x - 1} \).

We can simplify this by canceling one \( (x - 1) \) from the numerator and denominator, leading to:

\( x(x - 1) \).

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

\( 1(1 - 1) = 1 \cdot 0 = 0 \).

Therefore, the limit of the function as \( x \) approaches 1 is 0.

In the second example, we find the limit of the same function as \( x \) approaches 2. Here, substituting \( x = 2 \) into the denominator gives us \( 2 - 1 = 1 \), which is not zero. This allows us to directly substitute \( x = 2 \) into the function:

\( \frac{2^3 - 2 \cdot 2^2 + 2}{2 - 1} \).

Calculating the numerator, we have:

\( 2^3 = 8 \), \( 2 \cdot 2^2 = 8 \), so \( 8 - 8 + 2 = 2 \).

Thus, the limit simplifies to:

\( \frac{2}{1} = 2 \).

In summary, when finding limits of rational functions, if the denominator approaches zero, factor the expression to cancel common terms before substituting. If the denominator does not approach zero, direct substitution is sufficient. This understanding is crucial for solving limits effectively in calculus.