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→∞f(x)=L\lim_{x \to \infty} f(x) = L

As x approaches negative infinity: lim⁡x→−∞f(x)=L\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→∞(3x4−5x2+7)=lim⁡x→∞3x4=∞\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→∞3x2+5x−12x2−7\lim_{x \to \infty} \frac{3x^2 + 5x - 1}{2x^2 - 7}

Step 1: Divide every term by x2x^2

=lim⁡x→∞3x2x2+5xx2−1x22x2x2−7x2= \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→∞3+5x−1x22−7x2= \lim_{x \to \infty} \frac{3 + \frac{5}{x} - \frac{1}{x^2}}{2 - \frac{7}{x^2}}

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

=3+0−02−0=32= \frac{3 + 0 - 0}{2 - 0} = \frac{3}{2}

The Key Insight

lim⁡x→∞1x=0\lim_{x \to \infty} \frac{1}{x} = 0 lim⁡x→∞1xn=0 for any n>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→∞f(x)=L\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 anxn+...bmxm+...\frac{a_nx^n + ...}{b_mx^m + ...}:

  1. n<mn < m (denominator degree higher): Limit = 0
  2. n=mn = m (same degree): Limit = anbm\frac{a_n}{b_m} (ratio of leading coefficients)
  3. n>mn > m (numerator degree higher): Limit = ±∞\pm\infty

Example

lim⁡x→∞5x3+2xx3−4\lim_{x \to \infty} \frac{5x^3 + 2x}{x^3 - 4}

Same degree (both 3), so:

Limit=51=5\text{Limit} = \frac{5}{1} = 5

Horizontal asymptote at y = 5!

📚 Practice Problems

1Problem 1medium

❓ Question:

Evaluate lim⁡x→∞4x2−3x+12x2+5\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→∞4x2−3x+12x2+5=42=2\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 x2x^2:

lim⁡x→∞4−3x+1x22+5x2=4−0+02+0=42=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→∞3x2−5x+12x2+x−4\lim_{x \to \infty} \frac{3x^2 - 5x + 1}{2x^2 + x - 4} b) lim⁡x→∞5x3+2xx2−1\lim_{x \to \infty} \frac{5x^3 + 2x}{x^2 - 1} c) lim⁡x→−∞4x−72x2+3\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 (x2x^2):

lim⁡x→∞3−5x+1x22+1x−4x2\lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{1}{x} - \frac{4}{x^2}}

As x→∞x \to \infty, terms with xx in denominator approach 0:

=3−0+02+0−0=32= \frac{3 - 0 + 0}{2 + 0 - 0} = \frac{3}{2}

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

Divide by x3x^3:

lim⁡x→∞5+2x21x−1x3\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 x2x^2:

lim⁡x→−∞4x−7x22+3x2\lim_{x \to -\infty} \frac{\frac{4}{x} - \frac{7}{x^2}}{2 + \frac{3}{x^2}}

Numerator approaches 0, denominator approaches 2:

=02=0= \frac{0}{2} = 0

3Problem 3medium

❓ Question:

Evaluate the following limits:

a) lim⁡x→∞3x2−5x+12x2+x−4\lim_{x \to \infty} \frac{3x^2 - 5x + 1}{2x^2 + x - 4} b) lim⁡x→∞5x3+2xx2−1\lim_{x \to \infty} \frac{5x^3 + 2x}{x^2 - 1} c) lim⁡x→−∞4x−72x2+3\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 (x2x^2):

lim⁡x→∞3−5x+1x22+1x−4x2\lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{1}{x} - \frac{4}{x^2}}

As x→∞x \to \infty, terms with xx in denominator approach 0:

=3−0+02+0−0=32= \frac{3 - 0 + 0}{2 + 0 - 0} = \frac{3}{2}

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

Divide by x3x^3:

lim⁡x→∞5+2x21x−1x3\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 x2x^2:

lim⁡x→−∞4x−7x22+3x2\lim_{x \to -\infty} \frac{\frac{4}{x} - \frac{7}{x^2}}{2 + \frac{3}{x^2}}

Numerator approaches 0, denominator approaches 2:

=02=0= \frac{0}{2} = 0

4Problem 4medium

❓ Question:

Evaluate lim⁡x→−∞7x3x2+1\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→−∞7x3x2+1=0\lim_{x \to -\infty} \frac{7x}{3x^2 + 1} = 0

Verification: Divide all terms by x2x^2:

lim⁡x→−∞7xx23x2x2+1x2=lim⁡x→−∞7x3+1x2\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→−∞x \to -\infty: 7x→0\frac{7}{x} \to 0 and 1x2→0\frac{1}{x^2} \to 0

=03+0=0= \frac{0}{3 + 0} = 0 ✓

Answer: 0

5Problem 5hard

❓ Question:

Find lim(x→∞) (3x² - 5x + 2)/(x² + 4x - 1)

💡 Show Solution

Step 1: Identify highest power in denominator: Highest power is x²

Step 2: Divide every term by x²: [(3x²/x²) - (5x/x²) + (2/x²)]/[(x²/x²) + (4x/x²) - (1/x²)] = [3 - 5/x + 2/x²]/[1 + 4/x - 1/x²]

Step 3: Evaluate as x→∞: As x→∞: 1/x→0, 1/x²→0

Step 4: Substitute limits: [3 - 0 + 0]/[1 + 0 - 0] = 3/1 = 3

Step 5: Shortcut rule: When degrees are equal, limit = ratio of leading coefficients 3x²/x² = 3

Answer: 3

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