To find the limit of the function $f(x) = \frac{x^2 - 3x + 2}{x^2 + x - 6}$, you need to analyze the function's behavior as $x$ approaches a specific value.

Let's break it down step by step.

Step 1: Factor the numerator and denominator

First, factor both the numerator and the denominator of the function.

• The numerator is $x^2 - 3x + 2$. This can be factored as: $x^2 - 3x + 2 = (x - 1)(x - 2)$
• The denominator is $x^2 + x - 6$. This can be factored as: $x^2 + x - 6 = (x - 2)(x + 3)$

So, the function can be rewritten as: $f(x) = \frac{(x - 1)(x - 2)}{(x - 2)(x + 3)}$

Step 2: Simplify the expression

Notice that the factor $(x - 2)$ appears in both the numerator and the denominator. Assuming $x \neq 2$, these terms can be canceled out: $f(x) = \frac{x - 1}{x + 3}$ However, it's important to remember that the original function is undefined at $x = 2$ because the denominator becomes zero there.

Step 3: Find the limit

Now, you can find the limit as $x$ approaches a specific value. Let's consider two common cases:

Case 1: $\lim_{x \to 2} f(x)$

Since the function is undefined at $x = 2$, we instead find the limit of the simplified expression: $\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x - 1}{x + 3} = \frac{2 - 1}{2 + 3} = \frac{1}{5}$

Case 2: $\lim_{x \to \infty} f(x)$

For very large values of $x$, the behavior of the function is dominated by the highest degree terms in the numerator and the denominator. The highest degree terms are $x$ in the numerator and $x$ in the denominator: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x - 1}{x + 3} = \lim_{x \to \infty} \frac{x(1 - \frac{1}{x})}{x(1 + \frac{3}{x})} = \frac{1}{1} = 1$

Summary of Limits

• $\lim_{x \to 2} f(x) = \frac{1}{5}$
• $\lim_{x \to \infty} f(x) = 1$

Would you like details on any of these steps, or do you have any other questions?

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Here are 5 related questions to explore:

1. How do you handle limits when the function is undefined at a certain point?
2. What is the importance of factoring in finding limits?
3. How do you approach limits that involve infinity?
4. What is a removable discontinuity in the context of limits?
5. How can limits be used to find the derivative of a function?

Tip: Always consider simplifying the function before applying limit rules, especially if the function is undefined at a point. This can often reveal removable discontinuities.