Mastering Limits A Step-by-Step Solution Of Lim (x->2) [1/(x-2) - (2(2x-3))/(x^3-3x^2+2x)]

Aug 2, 2025 by Kenji Nakamura 91 views

$Decoding the Limit: A Deep Dive into ${\lim _{x \rightarrow 2}\left(\frac{1}{x-2}-\frac{2(2 x-3)}{x^3-3 x^2+2 x}\right)}$$

Hey everyone! Today, we're going to break down a fascinating limit problem that might seem a bit intimidating at first glance. We're tackling:

\lim _{x \rightarrow 2}\left(\frac{1}{x-2}-\frac{2(2 x-3)}{x^3-3 x^2+2 x}\right)

This isn't just about plugging in numbers; it's about understanding how functions behave as they approach a specific point. Limits are a cornerstone of calculus, and mastering them opens doors to understanding continuity, derivatives, and integrals. So, let's dive in and make this limit crystal clear!

Unraveling the Limit

When we first encounter a limit problem, especially one like this with fractions and polynomials, the initial instinct might be to directly substitute the value that x is approaching. However, if we try to plug in x = 2 directly into our expression, we quickly run into a problem. The denominator of the first fraction, (x - 2), becomes zero, and the denominator of the second fraction, x³ - 3x² + 2x, also evaluates to zero. This gives us an indeterminate form, which means we can't determine the limit simply by substitution. We need to do some algebraic manipulation first.

The key to solving this type of limit is to simplify the expression. This often involves factoring, combining fractions, and canceling out common factors. Our goal is to rewrite the expression in a form where we can directly substitute x = 2 without encountering a zero in the denominator.

Let's start by factoring the cubic polynomial in the denominator of the second fraction. We have x³ - 3x² + 2x. Notice that x is a common factor, so we can factor it out: x(x² - 3x + 2). Now we have a quadratic expression inside the parentheses. We can further factor this quadratic into (x - 2)(x - 1). So, the fully factored denominator is x(x - 2)(x - 1). Factoring is crucial because it often reveals common factors that can be canceled, simplifying the expression and eliminating the indeterminate form. By factoring, we transform a complex polynomial into a product of simpler terms, making it easier to see potential cancellations and understand the behavior of the function.

Rewriting our original expression with the factored denominator, we get:

\lim _{x \rightarrow 2}\left(\frac{1}{x-2}-\frac{2(2 x-3)}{x(x-2)(x-1)}\right)

Now we can see a common factor of (x - 2) in both denominators, which is a good sign! To combine the fractions, we need a common denominator. The least common denominator (LCD) for the two fractions is x(x - 2)(x - 1). We'll rewrite the first fraction with this denominator by multiplying both the numerator and denominator by x(x - 1). This gives us:

\lim _{x \rightarrow 2}\left(\frac{x(x-1)}{x(x-2)(x-1)}-\frac{2(2 x-3)}{x(x-2)(x-1)}\right)

Now that the fractions have a common denominator, we can combine them by subtracting the numerators:

\lim _{x \rightarrow 2}\left(\frac{x(x-1)-2(2 x-3)}{x(x-2)(x-1)}\right)

Next, we'll simplify the numerator by expanding and combining like terms. Expanding x(x - 1) gives us x² - x, and expanding -2(2x - 3) gives us -4x + 6. Combining these, we have:

\lim _{x \rightarrow 2}\left(\frac{x^2-x-4x+6}{x(x-2)(x-1)}\right) = \lim _{x \rightarrow 2}\left(\frac{x^2-5x+6}{x(x-2)(x-1)}\right)

Now we have a quadratic expression in the numerator. Let's factor it. We're looking for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. So, we can factor the numerator as (x - 2)(x - 3). Our expression now looks like this:

\lim _{x \rightarrow 2}\left(\frac{(x-2)(x-3)}{x(x-2)(x-1)}\right)

Here's the crucial step: we can now cancel the common factor of (x - 2) from the numerator and the denominator. This is the step that eliminates the indeterminate form we encountered initially. Canceling the (x - 2) term, we get:

\lim _{x \rightarrow 2}\left(\frac{x-3}{x(x-1)}\right)

Evaluating the Simplified Limit

We've done the hard work of simplifying the expression. Now, we can finally try direct substitution. Let's plug in x = 2 into the simplified expression:

\frac{2-3}{2(2-1)} = \frac{-1}{2(1)} = -\frac{1}{2}

And there we have it! The limit of the original expression as x approaches 2 is -1/2. This means that as x gets closer and closer to 2, the value of the function gets closer and closer to -1/2. The process of simplifying the expression through factoring and canceling common factors was essential to avoid the indeterminate form and arrive at the correct answer.

Key Takeaways and Strategies

This problem highlights several important concepts and strategies for tackling limits, particularly those involving rational functions (fractions with polynomials). Let's recap the key takeaways:

Direct Substitution First: Always try direct substitution first. If it works (i.e., doesn't result in an indeterminate form like 0/0), you're done! However, as we saw in this case, direct substitution often leads to an indeterminate form, signaling the need for further manipulation.
Indeterminate Forms: Recognize indeterminate forms like 0/0, ∞/∞, 0 * ∞, ∞ - ∞, 0⁰, 1^∞, and ∞⁰. These forms don't tell us anything about the limit directly and require algebraic manipulation or other techniques to resolve.
Factoring is Your Friend: Factoring polynomials is a powerful technique for simplifying expressions and revealing common factors that can be canceled. Look for common factors, difference of squares, quadratic factoring, and other factoring patterns.
Common Denominators: When dealing with sums or differences of fractions, find a common denominator to combine the terms into a single fraction. This often simplifies the expression and makes it easier to manipulate.
Cancellation is Key: Canceling common factors from the numerator and denominator is often the crucial step in resolving indeterminate forms. This eliminates the source of the zero in the denominator, allowing for direct substitution.
Simplify Before Evaluating: The general strategy for limits is to simplify the expression as much as possible before attempting to evaluate the limit. This often involves algebraic manipulation, trigonometric identities, or other techniques specific to the problem.
Practice Makes Perfect: Like any mathematical skill, mastering limits requires practice. Work through a variety of problems, and you'll start to recognize patterns and develop a toolkit of techniques for tackling different types of limits.

Stepping Beyond: More Challenging Limits

Now that we've conquered this limit, let's briefly touch on some other types of limits and techniques you might encounter. This problem involved a rational function and algebraic manipulation. However, limits can involve trigonometric functions, exponential functions, logarithmic functions, and more complex expressions.

For trigonometric limits, you'll often use trigonometric identities and special limits like ${\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1}$ and ${\lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x} = 0}$ . These limits are fundamental building blocks for evaluating more complex trigonometric limits.

When dealing with limits involving infinity (e.g., ${\lim_{x \rightarrow \infty} \frac{1}{x}}$ ), you'll often consider the behavior of the function as x becomes very large. For rational functions, you can divide the numerator and denominator by the highest power of x to simplify the expression and determine the limit.

For more complex limits, you might need to use L'Hôpital's Rule, which applies to indeterminate forms of the type 0/0 or ∞/∞. L'Hôpital's Rule states that if the limit of the ratio of the derivatives exists, then the limit of the original ratio also exists and is equal to the limit of the ratio of the derivatives. This rule can be a powerful tool for evaluating limits that are difficult to solve using other methods.

Finally, remember that limits are not just abstract mathematical concepts. They have real-world applications in physics, engineering, economics, and other fields. Understanding limits is essential for modeling continuous change and understanding the behavior of functions in various contexts.

So, keep practicing, keep exploring, and keep pushing your understanding of limits! They are a fundamental concept in calculus and a gateway to many exciting mathematical ideas. Good luck, and have fun with it!