How To Evaluate A Limit At Infinity

Limits at infinity, a fundamental concept in calculus, allow us to analyze the behavior of functions as their input values grow without bound. Mastering the techniques for evaluating these limits is crucial for understanding asymptotes, analyzing the long-term behavior of mathematical models, and grasping more advanced calculus concepts. This article delves into the methods for evaluating limits at infinity, providing a comprehensive guide with examples and explanations.

Understanding Limits at Infinity

Before diving into the techniques, let's define what we mean by a limit at infinity. When we write lim x→∞ f(x) = L, it means that as x becomes arbitrarily large (approaches infinity), the value of the function f(x) gets arbitrarily close to L. Similarly, lim x→−∞ f(x) = L means that as x becomes arbitrarily large in the negative direction (approaches negative infinity), the value of the function f(x) approaches L. Note that L can be a finite number, infinity (∞), or negative infinity (−∞).

Techniques for Evaluating Limits at Infinity

Several techniques can be used to evaluate limits at infinity, each suited for different types of functions. We'll explore these methods in detail:

1. Direct Substitution

The simplest approach is direct substitution. However, this method only works if the function is continuous at infinity, which is rarely the case. If direct substitution results in a determinate form (like a finite number), then that's the limit. However, if it results in an indeterminate form (like ∞/∞ or ∞ − ∞), we need to use other techniques.

2. Dividing by the Highest Power of x

This is a powerful technique for rational functions (ratios of polynomials). The idea is to divide both the numerator and denominator by the highest power of x that appears in the denominator. This simplifies the expression and allows us to evaluate the limit more easily.

Example 1:

Evaluate lim x→∞ (3x² + 2x + 1) / (4x² - x + 2)

• Step 1: Identify the highest power of x in the denominator. In this case, it's x².

• Step 2: Divide both the numerator and denominator by x².

lim x→∞ (3x² + 2x + 1) / (4x² - x + 2) = lim x→∞ (3 + 2/x + 1/x²) / (4 - 1/x + 2/x²)

• Step 3: Evaluate the limit as x approaches infinity. Remember that lim x→∞ (1/xⁿ) = 0 for any positive integer n.

lim x→∞ (3 + 2/x + 1/x²) / (4 - 1/x + 2/x²) = (3 + 0 + 0) / (4 - 0 + 0) = 3/4

Therefore, lim x→∞ (3x² + 2x + 1) / (4x² - x + 2) = 3/4.

Example 2:

Evaluate lim x→∞ (x + 1) / (x³)

• Step 1: Identify the highest power of x in the denominator. In this case, it's x³.

• Step 2: Divide both the numerator and denominator by x³.

lim x→∞ (x + 1) / (x³) = lim x→∞ (1/x² + 1/x³) / 1

• Step 3: Evaluate the limit as x approaches infinity.

lim x→∞ (1/x² + 1/x³) / 1 = (0 + 0) / 1 = 0

Therefore, lim x→∞ (x + 1) / (x³) = 0.

Example 3:

Evaluate lim x→∞ (x³) / (x + 1)

• Step 1: Identify the highest power of x in the denominator. In this case, it's x.

• Step 2: Divide both the numerator and denominator by x.

lim x→∞ (x³) / (x + 1) = lim x→∞ (x²) / (1 + 1/x)

• Step 3: Evaluate the limit as x approaches infinity.

lim x→∞ (x²) / (1 + 1/x) = ∞ / (1 + 0) = ∞

Therefore, lim x→∞ (x³) / (x + 1) = ∞.

3. Conjugate Multiplication

This technique is useful when dealing with expressions involving square roots. Multiplying by the conjugate can help eliminate the square root and simplify the expression.

Example:

Evaluate lim x→∞ (√(x² + x) - x)

• Step 1: Multiply the expression by its conjugate, (√(x² + x) + x) / (√(x² + x) + x).

lim x→∞ (√(x² + x) - x) * (√(x² + x) + x) / (√(x² + x) + x)

• Step 2: Simplify the numerator using the difference of squares formula (a - b)(a + b) = a² - b².

lim x→∞ (x² + x - x²) / (√(x² + x) + x) = lim x→∞ x / (√(x² + x) + x)

• Step 3: Divide both the numerator and denominator by x (which is equivalent to √(x²) inside the square root).

lim x→∞ x / (√(x² + x) + x) = lim x→∞ 1 / (√(1 + 1/x) + 1)

• Step 4: Evaluate the limit as x approaches infinity.

lim x→∞ 1 / (√(1 + 1/x) + 1) = 1 / (√(1 + 0) + 1) = 1 / (1 + 1) = 1/2

Therefore, lim x→∞ (√(x² + x) - x) = 1/2.

4. L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for evaluating limits of indeterminate forms such as 0/0 or ∞/∞. It states that if lim x→c f(x) / g(x) is of the form 0/0 or ∞/∞, and if f'(x) and g'(x) exist and g'(x) ≠ 0, then:

lim x→c f(x) / g(x) = lim x→c f'(x) / g'(x)

This means we can take the derivative of the numerator and the derivative of the denominator separately and then evaluate the limit again. We can apply L'Hôpital's Rule repeatedly as long as the limit remains in an indeterminate form. It's crucial to verify that the limit is indeed an indeterminate form before applying L'Hôpital's Rule.

Example 1:

Evaluate lim x→∞ x / eˣ

• Step 1: Verify that the limit is of the form ∞/∞. As x approaches infinity, both x and eˣ approach infinity.

• Step 2: Apply L'Hôpital's Rule.

lim x→∞ x / eˣ = lim x→∞ 1 / eˣ

• Step 3: Evaluate the limit as x approaches infinity.

lim x→∞ 1 / eˣ = 0

Therefore, lim x→∞ x / eˣ = 0.

Example 2:

Evaluate lim x→∞ ln(x) / x

• Step 1: Verify that the limit is of the form ∞/∞. As x approaches infinity, both ln(x) and x approach infinity.

• Step 2: Apply L'Hôpital's Rule.

lim x→∞ ln(x) / x = lim x→∞ (1/x) / 1 = lim x→∞ 1/x

• Step 3: Evaluate the limit as x approaches infinity.

lim x→∞ 1/x = 0

Therefore, lim x→∞ ln(x) / x = 0.

Important Note: L'Hôpital's Rule should only be applied when the limit is in an indeterminate form (0/0 or ∞/∞). Applying it to other forms will lead to incorrect results.

5. Squeeze Theorem (Sandwich Theorem)

The Squeeze Theorem states that if we have three functions, f(x), g(x), and h(x), such that f(x) ≤ g(x) ≤ h(x) for all x in an interval containing c (except possibly at c itself), and if lim x→c f(x) = L and lim x→c h(x) = L, then lim x→c g(x) = L.

While less commonly used for limits at infinity directly, the Squeeze Theorem can be helpful when dealing with oscillating functions or functions bounded by other simpler functions.

Example:

Evaluate lim x→∞ (sin x) / x

• Step 1: We know that -1 ≤ sin x ≤ 1 for all x.

• Step 2: Divide all parts of the inequality by x (assuming x > 0, which is true as x approaches infinity).

-1/x ≤ (sin x) / x ≤ 1/x

• Step 3: Evaluate the limits of the bounding functions as x approaches infinity.

lim x→∞ (-1/x) = 0 and lim x→∞ (1/x) = 0

• Step 4: By the Squeeze Theorem, since -1/x ≤ (sin x) / x ≤ 1/x and both -1/x and 1/x approach 0 as x approaches infinity, then

lim x→∞ (sin x) / x = 0

Therefore, lim x→∞ (sin x) / x = 0.

6. Recognizing Dominant Terms

In some cases, we can evaluate limits at infinity by recognizing the dominant terms in an expression. A dominant term is the term that grows the fastest as x approaches infinity. For example, in a polynomial, the term with the highest power of x is the dominant term.

Example:

Evaluate lim x→∞ (5x³ + 2x² - 7) / (2x³ - x + 4)

• Step 1: Identify the dominant terms in the numerator and denominator. In this case, they are 5x³ and 2x³, respectively.

• Step 2: Divide both the numerator and denominator by the dominant term, x³.

lim x→∞ (5x³ + 2x² - 7) / (2x³ - x + 4) = lim x→∞ (5 + 2/x - 7/x³) / (2 - 1/x² + 4/x³)

• Step 3: Evaluate the limit as x approaches infinity.

lim x→∞ (5 + 2/x - 7/x³) / (2 - 1/x² + 4/x³) = (5 + 0 - 0) / (2 - 0 + 0) = 5/2

Therefore, lim x→∞ (5x³ + 2x² - 7) / (2x³ - x + 4) = 5/2. This technique is essentially a shortcut of dividing by the highest power in the denominator, focusing solely on the terms that dictate the overall behavior.

Limits at Negative Infinity

The techniques used for evaluating limits at positive infinity can also be applied to limits at negative infinity. However, there are a few additional considerations:

• Even and Odd Functions: Remember the behavior of even and odd functions. For an even function, f(x) = f(-x), so lim x→∞ f(x) = lim x→−∞ f(x). For an odd function, f(x) = -f(-x), so lim x→∞ f(x) = - lim x→−∞ f(x).

• Sign Changes with Radicals: When dealing with square roots or other even roots, be careful with signs when substituting -x for x. For example, √(x²) = |x|, which is x if x is positive and -x if x is negative.

Example:

Evaluate lim x→−∞ (√(x² + 1) + x)

• Step 1: Factor out x² from under the square root. Since we're approaching negative infinity, we need to be careful about the sign of √(x²).

lim x→−∞ (√(x² + 1) + x) = lim x→−∞ (√[x²(1 + 1/x²)] + x) = lim x→−∞ (|x|√(1 + 1/x²) + x)

• Step 2: Since x is approaching negative infinity, |x| = -x.

lim x→−∞ (-x√(1 + 1/x²) + x)

• Step 3: Factor out x.

lim x→−∞ x(-√(1 + 1/x²) + 1)

• Step 4: Evaluate the limit as x approaches negative infinity.

lim x→−∞ x(-√(1 + 1/x²) + 1) = -∞(-√(1 + 0) + 1) = -∞(-1 + 1) = -∞(0)

This is an indeterminate form. We need to use a different approach. Let's multiply by the conjugate:

lim x→−∞ (√(x² + 1) + x) * (√(x² + 1) - x) / (√(x² + 1) - x) = lim x→−∞ (x² + 1 - x²) / (√(x² + 1) - x) = lim x→−∞ 1 / (√(x² + 1) - x)

Now, factor out x² from under the square root, remembering |x| = -x:

lim x→−∞ 1 / (√(x² + 1) - x) = lim x→−∞ 1 / (-x√(1 + 1/x²) - x) = lim x→−∞ 1 / [x(-√(1 + 1/x²) - 1)]

Evaluate the limit:

lim x→−∞ 1 / [x(-√(1 + 1/x²) - 1)] = 1 / [-∞(-√(1 + 0) - 1)] = 1 / [-∞(-1 - 1)] = 1 / [-∞(-2)] = 1 / ∞ = 0

Therefore, lim x→−∞ (√(x² + 1) + x) = 0.

Common Mistakes to Avoid

• Incorrectly Applying L'Hôpital's Rule: Make sure the limit is in an indeterminate form (0/0 or ∞/∞) before applying L'Hôpital's Rule.
• Ignoring Signs with Radicals: When dealing with limits at negative infinity and even roots, pay close attention to the signs. Remember that √(x²) = |x|.
• Forgetting to Divide by the Highest Power in the Denominator: When using the method of dividing by the highest power of x, always divide by the highest power in the denominator, not the numerator.
• Assuming All Functions Approach Zero at Infinity: Only functions of the form 1/xⁿ (where n is a positive number) approach zero as x approaches infinity. Functions like ln(x), x, and eˣ approach infinity.
• Not Simplifying Before Evaluating: Always simplify the expression as much as possible before attempting to evaluate the limit. This often involves factoring, combining terms, or multiplying by the conjugate.

Practical Applications of Limits at Infinity

Limits at infinity have many practical applications in various fields:

• Physics: Analyzing the long-term behavior of physical systems. For example, determining the terminal velocity of an object falling through the air involves evaluating a limit at infinity.
• Engineering: Designing control systems and analyzing the stability of systems. Limits at infinity help determine if a system will remain stable over time.
• Economics: Modeling economic growth and predicting long-term trends.
• Computer Science: Analyzing the efficiency of algorithms. The "Big O" notation used to describe the time complexity of algorithms relies on the concept of limits at infinity.
• Calculus and Analysis: Limits at infinity are fundamental to defining asymptotes, improper integrals, and the convergence of series.

Conclusion

Evaluating limits at infinity is a crucial skill in calculus and beyond. By mastering the techniques discussed in this article – direct substitution, dividing by the highest power of x, conjugate multiplication, L'Hôpital's Rule, the Squeeze Theorem, and recognizing dominant terms – you'll be well-equipped to analyze the behavior of functions as their input values grow without bound. Remember to practice these techniques with various examples and be mindful of the common mistakes to avoid. Understanding limits at infinity opens the door to a deeper understanding of calculus and its many applications in the real world.