Limits at Infinity

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

1. $n < m$ (denominator degree higher): Limit = 0
2. $n = m$ (same degree): Limit = $\frac{a_n}{b_m}$ (ratio of leading coefficients)
3. $n > m$ (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!

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:

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

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