How To Evaluate Limits At Infinity
penangjazz
Nov 24, 2025 · 13 min read
Table of Contents
Limits at infinity are a fundamental concept in calculus, allowing us to understand the behavior of functions as their input values grow without bound. Evaluating these limits is essential for analyzing asymptotes, understanding the end behavior of functions, and solving various problems in physics, engineering, and economics. This comprehensive guide will provide you with a step-by-step approach to evaluating limits at infinity, along with examples and practical tips to master this skill.
Understanding Limits at Infinity
The concept of a limit at infinity explores what happens to the value of a function, f(x), as x approaches either positive infinity (x → ∞) or negative infinity (x → -∞). It's not about x actually reaching infinity, but rather observing the trend of f(x) as x becomes arbitrarily large (positive or negative).
Formal Definition:
While a rigorous mathematical definition involves epsilon-delta arguments, the core idea is:
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The limit of f(x) as x approaches infinity is L (written as lim x→∞ f(x) = L) if the values of f(x) get arbitrarily close to L as x becomes sufficiently large.
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Similarly, the limit of f(x) as x approaches negative infinity is L (written as lim x→-∞ f(x) = L) if the values of f(x) get arbitrarily close to L as x becomes sufficiently negative.
Why are Limits at Infinity Important?
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Asymptotes: Limits at infinity help identify horizontal asymptotes of a function's graph. If lim x→∞ f(x) = L or lim x→-∞ f(x) = L, then y = L is a horizontal asymptote.
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End Behavior: They describe how a function behaves for very large positive or negative input values. This is crucial in modeling real-world phenomena over extended periods or vast scales.
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Calculus Foundations: Limits at infinity are essential for understanding improper integrals, series convergence, and other advanced calculus topics.
Strategies for Evaluating Limits at Infinity
Evaluating limits at infinity requires different techniques depending on the type of function. Here are some common strategies:
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Rational Functions (Polynomials Divided by Polynomials): This is the most common type encountered.
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Functions with Radicals: Involves manipulating the expression to handle square roots, cube roots, etc.
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Exponential and Logarithmic Functions: Requires knowledge of their asymptotic behavior.
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Trigonometric Functions: Often involves the Squeeze Theorem.
1. Evaluating Limits of Rational Functions at Infinity
This is perhaps the most fundamental technique. The key idea is to divide both the numerator and denominator by the highest power of x present in the denominator. This simplifies the expression and allows us to evaluate the limit more easily.
Steps:
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Identify the Highest Power of x in the Denominator: Let's call this x<sup>n</sup>.
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Divide Numerator and Denominator by x<sup>n</sup>: This step is crucial for simplifying the expression.
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Simplify: Divide each term in the numerator and denominator by x<sup>n</sup>.
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Evaluate the Limit: As x approaches infinity, terms of the form c/x<sup>k</sup> (where c is a constant and k > 0) will approach 0.
Examples:
Example 1: Evaluate lim x→∞ (3x<sup>2</sup> + 2x - 1) / (2x<sup>2</sup> - x + 3)
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Highest Power in Denominator: x<sup>2</sup>
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Divide by x<sup>2</sup>: lim x→∞ ( (3x<sup>2</sup> + 2x - 1) / x<sup>2</sup> ) / ( (2x<sup>2</sup> - x + 3) / x<sup>2</sup> )
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Simplify: lim x→∞ (3 + 2/x - 1/x<sup>2</sup>) / (2 - 1/x + 3/x<sup>2</sup>)
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Evaluate: As x→∞, 2/x → 0, 1/x<sup>2</sup> → 0, and 3/x<sup>2</sup> → 0. Therefore, the limit becomes (3 + 0 - 0) / (2 - 0 + 0) = 3/2
Example 2: Evaluate lim x→-∞ (4x + 5) / (x<sup>3</sup> - 7)
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Highest Power in Denominator: x<sup>3</sup>
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Divide by x<sup>3</sup>: lim x→-∞ ( (4x + 5) / x<sup>3</sup> ) / ( (x<sup>3</sup> - 7) / x<sup>3</sup> )
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Simplify: lim x→-∞ (4/x<sup>2</sup> + 5/x<sup>3</sup>) / (1 - 7/x<sup>3</sup>)
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Evaluate: As x→-∞, 4/x<sup>2</sup> → 0, 5/x<sup>3</sup> → 0, and 7/x<sup>3</sup> → 0. Therefore, the limit becomes (0 + 0) / (1 - 0) = 0
Example 3: Evaluate lim x→∞ (x<sup>4</sup> + 1) / (x<sup>2</sup> - 2x)
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Highest Power in Denominator: x<sup>2</sup>
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Divide by x<sup>2</sup>: lim x→∞ ( (x<sup>4</sup> + 1) / x<sup>2</sup> ) / ( (x<sup>2</sup> - 2x) / x<sup>2</sup> )
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Simplify: lim x→∞ (x<sup>2</sup> + 1/x<sup>2</sup>) / (1 - 2/x)
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Evaluate: As x→∞, 1/x<sup>2</sup> → 0 and 2/x → 0. Therefore, the limit becomes (x<sup>2</sup> + 0) / (1 - 0) = x<sup>2</sup>. As x→∞, x<sup>2</sup> → ∞. So, the limit is ∞. This means the function increases without bound as x approaches infinity.
Key Takeaway: If the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator, the limit will be either ∞ or -∞. If the degree of the numerator is less than the degree of the denominator, the limit will be 0. If the degrees are equal, the limit will be the ratio of the leading coefficients.
2. Evaluating Limits of Functions with Radicals at Infinity
When dealing with functions containing radicals (square roots, cube roots, etc.), we need to be careful with signs, especially when x approaches negative infinity.
Steps:
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Identify the Dominant Term: Determine the term with the highest power of x under the radical.
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Factor out the Dominant Term: Factor out the highest power of x from under the radical.
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Simplify: Use the properties of radicals to simplify the expression. Remember that √(x<sup>2</sup>) = |x|, which is x if x > 0 and -x if x < 0. This is crucial when taking the limit as x approaches negative infinity.
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Divide by the Appropriate Power of x: Divide both the numerator and denominator by the power of x that now appears outside the radical.
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Evaluate the Limit: Use the techniques for rational functions.
Examples:
Example 1: Evaluate lim x→∞ √(4x<sup>2</sup> + x) / (x - 1)
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Dominant Term: 4x<sup>2</sup> under the square root.
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Factor out x<sup>2</sup>: lim x→∞ √(x<sup>2</sup>(4 + 1/x)) / (x - 1)
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Simplify: lim x→∞ |x|√(4 + 1/x) / (x - 1). Since x→∞, x is positive, so |x| = x. Thus, lim x→∞ x√(4 + 1/x) / (x - 1)
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Divide by x: lim x→∞ √(4 + 1/x) / (1 - 1/x)
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Evaluate: As x→∞, 1/x → 0. Therefore, the limit becomes √(4 + 0) / (1 - 0) = 2/1 = 2
Example 2: Evaluate lim x→-∞ √(x<sup>2</sup> + 1) / (2x + 3)
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Dominant Term: x<sup>2</sup> under the square root.
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Factor out x<sup>2</sup>: lim x→-∞ √(x<sup>2</sup>(1 + 1/x<sup>2</sup>)) / (2x + 3)
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Simplify: lim x→-∞ |x|√(1 + 1/x<sup>2</sup>) / (2x + 3). Since x→-∞, x is negative, so |x| = -x. Thus, lim x→-∞ -x√(1 + 1/x<sup>2</sup>) / (2x + 3)
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Divide by x: lim x→-∞ -√(1 + 1/x<sup>2</sup>) / (2 + 3/x)
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Evaluate: As x→-∞, 1/x<sup>2</sup> → 0 and 3/x → 0. Therefore, the limit becomes -√(1 + 0) / (2 + 0) = -1/2
Key Takeaway: Remember to consider the sign of x when simplifying radicals. When x approaches negative infinity, |x| = -x. Failing to account for this will lead to incorrect answers.
3. Evaluating Limits of Exponential and Logarithmic Functions at Infinity
Exponential and logarithmic functions have distinct behaviors as x approaches infinity. Understanding these behaviors is crucial for evaluating their limits.
Exponential Functions:
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If a > 1, then lim x→∞ a<sup>x</sup> = ∞ and lim x→-∞ a<sup>x</sup> = 0. For example, lim x→∞ 2<sup>x</sup> = ∞ and lim x→-∞ 2<sup>x</sup> = 0.
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If 0 < a < 1, then lim x→∞ a<sup>x</sup> = 0 and lim x→-∞ a<sup>x</sup> = ∞. For example, lim x→∞ (1/2)<sup>x</sup> = 0 and lim x→-∞ (1/2)<sup>x</sup> = ∞.
Logarithmic Functions:
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lim x→∞ ln(x) = ∞. The natural logarithm grows without bound as x increases.
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Logarithmic functions are not defined for non-positive values, so lim x→-∞ ln(x) does not exist. However, we can consider lim x→0<sup>+</sup> ln(x) = -∞.
Examples:
Example 1: Evaluate lim x→∞ e<sup>-x</sup>
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Rewrite: e<sup>-x</sup> = 1/e<sup>x</sup>
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Evaluate: As x→∞, e<sup>x</sup> → ∞, so 1/e<sup>x</sup> → 0. Therefore, lim x→∞ e<sup>-x</sup> = 0.
Example 2: Evaluate lim x→∞ (e<sup>x</sup> - e<sup>-x</sup>) / (e<sup>x</sup> + e<sup>-x</sup>)
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Divide by e<sup>x</sup>: lim x→∞ (1 - e<sup>-2x</sup>) / (1 + e<sup>-2x</sup>)
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Evaluate: As x→∞, e<sup>-2x</sup> → 0. Therefore, the limit becomes (1 - 0) / (1 + 0) = 1
Example 3: Evaluate lim x→∞ ln(x<sup>2</sup> + 1) / x
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This limit requires L'Hôpital's Rule (see below for more details). Taking the derivative of the numerator and denominator, we get: lim x→∞ (2x / (x<sup>2</sup> + 1)) / 1 = lim x→∞ 2x / (x<sup>2</sup> + 1)
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Now divide by x<sup>2</sup>: lim x→∞ (2/x) / (1 + 1/x<sup>2</sup>)
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Evaluate: As x→∞, 2/x → 0 and 1/x<sup>2</sup> → 0. Therefore, the limit becomes 0 / (1 + 0) = 0
Key Takeaway: Understand the basic behaviors of exponential and logarithmic functions. When dealing with combinations, manipulate the expression to isolate the exponential or logarithmic term.
4. Evaluating Limits of Trigonometric Functions at Infinity
Trigonometric functions like sine and cosine oscillate between -1 and 1. Therefore, their limits as x approaches infinity generally do not exist. However, we can use the Squeeze Theorem to evaluate limits involving trigonometric functions multiplied by other functions.
Squeeze Theorem (Sandwich Theorem):
If g(x) ≤ f(x) ≤ h(x) for all x in an interval containing c (except possibly at c itself), and lim x→c g(x) = L and lim x→c h(x) = L, then lim x→c f(x) = L.
Steps:
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Identify the Bounded Trigonometric Function: Look for sin(x), cos(x), or similar functions, as they are bounded between -1 and 1.
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Establish Inequalities: Use the fact that -1 ≤ sin(x) ≤ 1 and -1 ≤ cos(x) ≤ 1 to create inequalities.
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Multiply by the Other Function: Multiply all parts of the inequality by the function multiplying the trigonometric function.
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Evaluate the Limits of the Outer Functions: Evaluate the limits of the functions on the left and right sides of the inequality as x approaches infinity.
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Apply the Squeeze Theorem: If the limits of the outer functions are equal, then the limit of the original function exists and is equal to that value.
Examples:
Example 1: Evaluate lim x→∞ sin(x) / x
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Bounded Function: sin(x)
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Inequalities: -1 ≤ sin(x) ≤ 1
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Multiply by 1/x: -1/x ≤ sin(x) / x ≤ 1/x
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Evaluate Limits: lim x→∞ -1/x = 0 and lim x→∞ 1/x = 0
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Squeeze Theorem: Since lim x→∞ -1/x = lim x→∞ 1/x = 0, then lim x→∞ sin(x) / x = 0
Example 2: Evaluate lim x→∞ x * cos(1/x)
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Rewrite: Let u = 1/x. As x approaches infinity, u approaches 0. So the limit becomes lim u→0 (1/u) * cos(u) = lim u→0 cos(u) / u.
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This limit does not exist. As u approaches 0, cos(u) approaches 1, while u approaches 0. The limit is of the form 1/0, which is unbounded.
Key Takeaway: The Squeeze Theorem is your primary tool for evaluating limits of trigonometric functions at infinity. Remember to establish the correct inequalities and carefully evaluate the limits of the bounding functions.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool for evaluating limits of indeterminate forms, such as 0/0 or ∞/∞. While not always necessary for limits at infinity, it can be very helpful in certain cases.
L'Hôpital's Rule:
If lim x→c f(x) / g(x) is of the form 0/0 or ∞/∞, and f'(x) and g'(x) exist and g'(x) ≠ 0 near c, then:
lim x→c *f(x) / g(x) = lim x→c f'(x) / g'(x)
Important Notes:
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L'Hôpital's Rule only applies to indeterminate forms. You must verify that the limit is of the form 0/0 or ∞/∞ before applying the rule.
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You may need to apply L'Hôpital's Rule multiple times if the first application still results in an indeterminate form.
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L'Hôpital's Rule can sometimes make the problem more complicated, so consider other techniques first.
Examples:
Example 1: Evaluate lim x→∞ x / e<sup>x</sup>
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Indeterminate Form: ∞/∞
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Apply L'Hôpital's Rule: lim x→∞ 1 / e<sup>x</sup>
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Evaluate: As x→∞, e<sup>x</sup> → ∞, so 1/e<sup>x</sup> → 0. Therefore, lim x→∞ x / e<sup>x</sup> = 0
Example 2: Evaluate lim x→∞ ln(x) / x
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Indeterminate Form: ∞/∞
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Apply L'Hôpital's Rule: lim x→∞ (1/x) / 1
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Evaluate: As x→∞, 1/x → 0. Therefore, lim x→∞ ln(x) / x = 0
Key Takeaway: L'Hôpital's Rule is a valuable tool for indeterminate forms, but use it judiciously. Always check that the limit is of the correct form before applying the rule.
Common Mistakes to Avoid
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Ignoring Signs with Radicals: When simplifying expressions with radicals and x approaches negative infinity, remember that |x| = -x.
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Incorrectly Applying L'Hôpital's Rule: Only apply L'Hôpital's Rule to indeterminate forms (0/0 or ∞/∞).
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Not Simplifying Expressions: Always simplify the expression as much as possible before evaluating the limit.
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Forgetting Basic Limit Properties: Remember that lim x→∞ c/x<sup>k</sup> = 0 (where c is a constant and k > 0).
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Assuming Limits Always Exist: Not all functions have limits at infinity. Trigonometric functions, for example, often oscillate and do not approach a specific value.
Practice Problems
To solidify your understanding, try these practice problems:
- lim x→∞ (5x<sup>3</sup> - 2x + 1) / (2x<sup>3</sup> + x<sup>2</sup> - 4)
- lim x→-∞ (√(9x<sup>2</sup> - 5)) / (x + 2)
- lim x→∞ e<sup>2x</sup> / x<sup>2</sup>
- lim x→∞ (sin(x) + x) / x
- lim x→∞ ln(x) / √x
Conclusion
Evaluating limits at infinity is a crucial skill in calculus and its applications. By mastering the techniques discussed in this guide – including dividing by the highest power, handling radicals carefully, understanding exponential and logarithmic behaviors, and applying the Squeeze Theorem and L'Hôpital's Rule when appropriate – you'll be well-equipped to analyze the end behavior of functions and solve a wide range of problems. Remember to practice regularly and pay attention to common pitfalls to develop your proficiency. With consistent effort, you can confidently tackle limits at infinity and unlock a deeper understanding of calculus.
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