Evaluating Limits: Step-by-Step Solutions

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Evaluating Limits: Step-by-Step Solutions

Hey everyone! In this article, we're going to dive into evaluating limits, specifically focusing on two examples. We'll break down each step, so you can clearly understand how to arrive at the solution. Limits are a fundamental concept in calculus, and mastering them is crucial for further studies in mathematics. So, let's jump right in and tackle these problems together!

Evaluating lim x→0 3x/(x^2+6x-4)

Let's begin by examining the first limit: lim x→0 3x/(x^2+6x-4). When evaluating limits, our initial approach should always be direct substitution. This means we substitute the value that x is approaching (in this case, 0) directly into the expression. This helps us quickly determine if the limit can be found by simply plugging in the value. So, what happens when we do that?

Direct Substitution:

Substituting x = 0 into the expression, we get:

3(0) / (0^2 + 6(0) - 4) = 0 / -4 = 0

Guess what? We got a straightforward result! In this scenario, direct substitution worked perfectly. This indicates that the function is continuous at x = 0, meaning there are no surprises like division by zero or indeterminate forms. Therefore, the limit as x approaches 0 for the function 3x/(x^2+6x-4) is simply 0. Isn't that neat when the solution just pops out like that?

Why Direct Substitution Works (Sometimes):

The reason direct substitution works in some cases is because of the properties of continuous functions. A function is continuous at a point if there are no breaks, jumps, or holes in its graph at that point. For continuous functions, the limit as x approaches a value is simply the function's value at that point. Polynomials and rational functions (like the one we just looked at) are continuous everywhere except where the denominator of the rational function is zero. In our case, the denominator (x^2 + 6x - 4) is not zero when x = 0, so we can safely use direct substitution.

Key Takeaway:

Always start by trying direct substitution! It's the quickest way to find the limit if it works. If you get a determinate form (a real number), you've found your answer. However, if you encounter an indeterminate form (like 0/0), you'll need to employ other techniques, which we'll explore in the next example.

Evaluating lim x→2 (2x^(2 )+1)/(x^2+6x-4)

Now, let's tackle the second limit: lim x→2 (2x^(2 )+1)/(x^2+6x-4). Just like before, our first move should be to try direct substitution. This is our go-to strategy for evaluating limits, as it often provides a quick solution if the function is continuous at the point we're approaching. So, let's substitute x = 2 into the expression and see what happens.

Direct Substitution Attempt:

Substituting x = 2 into the expression, we get:

(2(2)^2 + 1) / ((2)^2 + 6(2) - 4) = (2(4) + 1) / (4 + 12 - 4) = (8 + 1) / (12) = 9 / 12

Look at that! We didn't encounter any indeterminate forms or division by zero. The result is a simple fraction, 9/12, which we can further simplify. Direct substitution worked like a charm in this case as well. This indicates that the function is continuous at x = 2, allowing us to directly substitute the value to find the limit.

Simplifying the Result:

We can simplify the fraction 9/12 by dividing both the numerator and the denominator by their greatest common divisor, which is 3:

9 / 12 = (9 ÷ 3) / (12 ÷ 3) = 3 / 4

So, the simplified result is 3/4. This is the value of the limit as x approaches 2 for the function (2x^(2 )+1)/(x^2+6x-4).

Why Direct Substitution Works Here Too:

Similar to the first example, direct substitution works in this case because we're dealing with a rational function, which is continuous everywhere except where the denominator is zero. When we substitute x = 2, the denominator (x^2 + 6x - 4) evaluates to 12, which is not zero. This confirms that the function is continuous at x = 2, making direct substitution a valid method for finding the limit.

Key Takeaway:

Direct substitution is a powerful tool for evaluating limits, especially for continuous functions like polynomials and rational functions. By directly substituting the value that x is approaching, we can often find the limit quickly and efficiently. Remember to always simplify your result if possible, as we did by reducing 9/12 to 3/4.

When Direct Substitution Fails: Indeterminate Forms

Okay, so we've seen how direct substitution can be super effective when evaluating limits. But what happens when it doesn't work? This is where things get a little more interesting! The most common scenario where direct substitution fails is when we encounter what's called an indeterminate form. Indeterminate forms are expressions that don't have a clear, defined value. The most common indeterminate forms you'll run into are 0/0 and ∞/∞, but there are others too.

What are Indeterminate Forms?

Think of indeterminate forms as mathematical puzzles. They arise when we try to evaluate a limit and end up with an expression that doesn't tell us anything definitive about the limit's value. For example, if we get 0/0, it doesn't necessarily mean the limit is 0 or infinity; it means we need to do more work to figure out what's going on. It's like a sign saying, "Hey, try a different approach!"

Why do Indeterminate Forms Happen?

Indeterminate forms often occur when there are competing factors in the limit that are pushing the expression in different directions. For example, in the case of 0/0, the numerator is approaching zero, which would make the fraction approach zero. But the denominator is also approaching zero, which would make the fraction approach infinity. These opposing forces create the indeterminacy.

What to do When You Encounter an Indeterminate Form:

When direct substitution leads to an indeterminate form, it's time to pull out some other techniques from your mathematical toolbox. Here are some common strategies:

  1. Algebraic Manipulation:

    • Factoring: If you have a rational function (a fraction with polynomials), try factoring the numerator and denominator. Sometimes, you can cancel out common factors that are causing the indeterminacy.
    • Rationalizing: If you have radicals (square roots, cube roots, etc.) in your expression, try rationalizing the numerator or denominator. This involves multiplying by a conjugate to eliminate the radical.
    • Simplifying Complex Fractions: If you have fractions within fractions, simplify them first.

  2. L'Hôpital's Rule:

    This is a powerful tool that can be used when you have an indeterminate form of 0/0 or ∞/∞. L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches a is indeterminate, then the limit is equal to the limit of f'(x)/g'(x) as x approaches a, where f'(x) and g'(x) are the derivatives of f(x) and g(x), respectively. In simpler terms, you take the derivative of the top and the derivative of the bottom and try the limit again. We'll see L'Hôpital's Rule in action in a later example!

  3. Trigonometric Identities:

    If your limit involves trigonometric functions, you might need to use trigonometric identities to simplify the expression.

  4. Squeeze Theorem:

    This theorem is useful when you can "squeeze" your function between two other functions whose limits are known.

Example of Factoring to Resolve an Indeterminate Form:

Let's say we have the limit: lim x→2 (x^2 - 4) / (x - 2)

If we try direct substitution, we get (2^2 - 4) / (2 - 2) = 0 / 0, which is an indeterminate form. So, we need to try a different approach. Notice that the numerator is a difference of squares, so we can factor it:

x^2 - 4 = (x - 2)(x + 2)

Now our limit looks like:

lim x→2 ((x - 2)(x + 2)) / (x - 2)

We can cancel the (x - 2) terms:

lim x→2 (x + 2)

Now, we can use direct substitution:

2 + 2 = 4

So, the limit is 4. By factoring, we were able to eliminate the indeterminate form and find the limit.

Key Takeaway:

Don't be discouraged when you encounter an indeterminate form! It just means you need to use a different technique. Indeterminate forms are a common challenge in evaluating limits, and mastering the techniques to resolve them is essential for success in calculus. Remember to try algebraic manipulation first, and if that doesn't work, consider L'Hôpital's Rule or other advanced techniques.

Conclusion

So, there you have it! We've walked through evaluating two limits step-by-step, highlighting the importance of direct substitution as a first approach. We've also discussed what to do when direct substitution doesn't work and introduces the concept of indeterminate forms and techniques to tackle them. Remember, practice makes perfect when it comes to limits. Keep working through examples, and you'll become a limit-evaluating pro in no time! Keep an eye out for more articles where we delve into more complex limit scenarios and techniques. Happy calculating, guys!