Limits at Infinity

Understanding what happens as x grows without bound

Limits at Infinity

What happens to a function as x gets really, really large? Or really, really negative?

The Notation

As x approaches positive infinity: $\lim_{x \to \infty} f(x) = L$

As x approaches negative infinity: $\lim_{x \to -\infty} f(x) = L$

These describe the end behavior of a function - where is it heading as we go far right or far left?

For Polynomials

The limit at infinity of a polynomial is determined by its leading term (highest power).

$\lim_{x \to \infty} (3x^4 - 5x^2 + 7) = \lim_{x \to \infty} 3x^4 = \infty$

Rule of Thumb:

Even degree, positive leading coefficient → $+\infty$ on both sides
Even degree, negative leading coefficient → $-\infty$ on both sides
Odd degree, positive leading coefficient → $-\infty$ (left), $+\infty$ (right)
Odd degree, negative leading coefficient → $+\infty$ (left), $-\infty$ (right)

For Rational Functions

With rational functions, divide everything by the highest power of x in the denominator:

$\lim_{x \to \infty} \frac{3x^2 + 5x - 1}{2x^2 - 7}$

Step 1: Divide every term by $x^2$

$= \lim_{x \to \infty} \frac{\frac{3x^2}{x^2} + \frac{5x}{x^2} - \frac{1}{x^2}}{\frac{2x^2}{x^2} - \frac{7}{x^2}}$

Step 2: Simplify

$= \lim_{x \to \infty} \frac{3 + \frac{5}{x} - \frac{1}{x^2}}{2 - \frac{7}{x^2}}$

Step 3: As $x \to \infty$ , terms like $\frac{1}{x} \to 0$

$= \frac{3 + 0 - 0}{2 - 0} = \frac{3}{2}$

The Key Insight

$\lim_{x \to \infty} \frac{1}{x} = 0$ $\lim_{x \to \infty} \frac{1}{x^n} = 0 \text{ for any } n > 0$

Big numbers make fractions tiny!

Horizontal Asymptotes

If $\lim_{x \to \infty} f(x) = L$ , then y = L is a horizontal asymptote.

The graph approaches this horizontal line as x gets large.

Three Cases for Rational Functions

For $\frac{a_nx^n + ...}{b_mx^m + ...}$ :

$n < m$ (denominator degree higher): Limit = 0
$n = m$ (same degree): Limit = $\frac{a_n}{b_m}$ (ratio of leading coefficients)
$n > m$ (numerator degree higher): Limit = $\pm\infty$

Example

$\lim_{x \to \infty} \frac{5x^3 + 2x}{x^3 - 4}$

Same degree (both 3), so:

$\text{Limit} = \frac{5}{1} = 5$

Horizontal asymptote at y = 5!

📚 Practice Problems

1Problem 1medium

❓ Question:

Evaluate $\lim_{x \to \infty} \frac{4x^2 - 3x + 1}{2x^2 + 5}$

💡 Show Solution

Step 1: Identify degrees

Numerator: degree 2 Denominator: degree 2

Same degree! The limit will be the ratio of leading coefficients.

Step 2: Find leading coefficients

Leading coefficient of numerator: 4 Leading coefficient of denominator: 2

Step 3: Take the ratio

$\lim_{x \to \infty} \frac{4x^2 - 3x + 1}{2x^2 + 5} = \frac{4}{2} = 2$

Answer: 2

Verification by division: Divide all terms by $x^2$ :

$\lim_{x \to \infty} \frac{4 - \frac{3}{x} + \frac{1}{x^2}}{2 + \frac{5}{x^2}} = \frac{4 - 0 + 0}{2 + 0} = \frac{4}{2} = 2$ ✓

2Problem 2medium

❓ Question:

Evaluate the following limits:

a) $\lim_{x \to \infty} \frac{3x^2 - 5x + 1}{2x^2 + x - 4}$ b) $\lim_{x \to \infty} \frac{5x^3 + 2x}{x^2 - 1}$ c) $\lim_{x \to -\infty} \frac{4x - 7}{2x^2 + 3}$

💡 Show Solution

Solution:

Part (a): Both numerator and denominator have degree 2.

Divide numerator and denominator by highest power ( $x^2$ ):

$\lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{1}{x} - \frac{4}{x^2}}$

As $x \to \infty$ , terms with $x$ in denominator approach 0:

$= \frac{3 - 0 + 0}{2 + 0 - 0} = \frac{3}{2}$

Part (b): Numerator degree (3) > denominator degree (2).

Divide by $x^3$ :

$\lim_{x \to \infty} \frac{5 + \frac{2}{x^2}}{\frac{1}{x} - \frac{1}{x^3}}$

Numerator approaches 5, denominator approaches 0 from positive side:

$= \infty$

Part (c): Numerator degree (1) < denominator degree (2).

Divide by $x^2$ :

$\lim_{x \to -\infty} \frac{\frac{4}{x} - \frac{7}{x^2}}{2 + \frac{3}{x^2}}$

Numerator approaches 0, denominator approaches 2:

$= \frac{0}{2} = 0$

3Problem 3medium

❓ Question:

Evaluate the following limits:

a) $\lim_{x \to \infty} \frac{3x^2 - 5x + 1}{2x^2 + x - 4}$ b) $\lim_{x \to \infty} \frac{5x^3 + 2x}{x^2 - 1}$ c) $\lim_{x \to -\infty} \frac{4x - 7}{2x^2 + 3}$

💡 Show Solution

Solution:

Part (a): Both numerator and denominator have degree 2.

Divide numerator and denominator by highest power ( $x^2$ ):

$\lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{1}{x} - \frac{4}{x^2}}$

As $x \to \infty$ , terms with $x$ in denominator approach 0:

$= \frac{3 - 0 + 0}{2 + 0 - 0} = \frac{3}{2}$

Part (b): Numerator degree (3) > denominator degree (2).

Divide by $x^3$ :

$\lim_{x \to \infty} \frac{5 + \frac{2}{x^2}}{\frac{1}{x} - \frac{1}{x^3}}$

Numerator approaches 5, denominator approaches 0 from positive side:

$= \infty$

Part (c): Numerator degree (1) < denominator degree (2).

Divide by $x^2$ :

$\lim_{x \to -\infty} \frac{\frac{4}{x} - \frac{7}{x^2}}{2 + \frac{3}{x^2}}$

Numerator approaches 0, denominator approaches 2:

$= \frac{0}{2} = 0$

4Problem 4medium

❓ Question:

Evaluate $\lim_{x \to -\infty} \frac{7x}{3x^2 + 1}$

💡 Show Solution

Step 1: Identify degrees

Numerator: degree 1 Denominator: degree 2

Denominator has higher degree!

Step 2: Apply the rule

When denominator degree > numerator degree, the limit is 0.

$\lim_{x \to -\infty} \frac{7x}{3x^2 + 1} = 0$

Verification: Divide all terms by $x^2$ :

$\lim_{x \to -\infty} \frac{\frac{7x}{x^2}}{\frac{3x^2}{x^2} + \frac{1}{x^2}} = \lim_{x \to -\infty} \frac{\frac{7}{x}}{3 + \frac{1}{x^2}}$

As $x \to -\infty$ : $\frac{7}{x} \to 0$ and $\frac{1}{x^2} \to 0$

$= \frac{0}{3 + 0} = 0$ ✓

Answer: 0

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