2.3 Limits Involving Infinity and Asymptotes

2.3 Limits Involving Infinity and Asymptotes

Core CLEP Concept: Asymptotic behavior describes how curves behave at their geometric boundaries—either exploding toward infinity near a vertical line or flattening out as $x$ grows without bound. On the CLEP Calculus exam, identifying vertical, horizontal, and slant asymptotes and calculating directional limits involving $\pm\infty$ are tested frequently.

---

1. Infinite Limits and Vertical Asymptotes

An infinite limit occurs when function values grow arbitrarily large (positive or negative) as $x$ approaches a finite number $c$.

Definition of a Vertical Asymptote

The vertical line $x = c$ is called a vertical asymptote of the curve $y = f(x)$ if at least one of the following statements is true:

$\lim_{x \to c^-} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to c^+} f(x) = \pm\infty$

Systematic Sign Analysis Near Vertical Asymptotes

When evaluating $\lim_{x \to c} \frac{P(x)}{Q(x)}$ where $P(c) \neq 0$ and $Q(c) = 0$:

1. The magnitude of the quotient approaches $\infty$.
2. Determine the sign ($+$ or $-$) by testing numbers slightly to the left ($c - 0.001$) and slightly to the right ($c + 0.001$).

Step-by-Step Worked Example: Vertical Asymptote Behavior

Problem: Analyze the one-sided limits of $f(x) = \frac{2 - x}{(x - 3)^2(x + 1)}$ at $x = 3$ and $x = -1$.

Analysis at $x = 3$:

• Numerator as $x \to 3$: $2 - 3 = -1$ (Negative).
• Denominator factor $(x - 3)^2$: Always positive for $x \neq 3$.
• Denominator factor $(x + 1)$: $3 + 1 = 4$ (Positive).
• Left-Hand Limit ($x \to 3^-$): $\frac{\text{Negative}}{(\text{Positive})(\text{Positive})} = -\infty$
• Right-Hand Limit ($x \to 3^+$): $\frac{\text{Negative}}{(\text{Positive})(\text{Positive})} = -\infty$
• Since both one-sided limits are $-\infty$, we write $\lim_{x \to 3} f(x) = -\infty$, and $x = 3$ is a vertical asymptote.

Analysis at $x = -1$:

• Numerator as $x \to -1$: $2 - (-1) = 3$ (Positive).
• Denominator factor $(x - 3)^2$: $(-1 - 3)^2 = 16$ (Positive).
• Denominator factor $(x + 1)$:
  • As $x \to -1^+$, $x > -1 \implies x + 1 > 0$ (Positive). Limit $= \frac{+}{(+)(+)} = +\infty$.
  • As $x \to -1^-$, $x < -1 \implies x + 1 < 0$ (Negative). Limit $= \frac{+}{(+)(-)} = -\infty$.
• Since the left and right behaviors differ, the two-sided limit DNE, but $x = -1$ remains a vertical asymptote.

---

2. Limits at Infinity and Horizontal Asymptotes

A limit at infinity investigates end behavior as $x \to +\infty$ (moving infinitely rightward) or $x \to -\infty$ (moving infinitely leftward).

Definition of a Horizontal Asymptote

The line $y = L$ is a horizontal asymptote of the curve $y = f(x)$ if:

$\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L$

(Important Distinction: A function can have at most two horizontal asymptotes—one for the positive direction and one for the negative direction. While a curve can never touch or cross a vertical asymptote at $x = c$, a curve can intersect its horizontal asymptote in the finite plane.)

---

3. Rational Functions at Infinity: Degree Comparison Theorem

Let $f(x) = \frac{a_n x^n + a_{n-1}x^{n-1} + \dots + a_0}{b_m x^m + b_{m-1}x^{m-1} + \dots + b_0}$ with $a_n, b_m \neq 0$.

Algebraic Method: Dividing by the Highest Denominator Power

To prove limits at infinity rigorously on the CLEP exam, divide every term in the numerator and denominator by $x^m$ (the highest power of $x$ appearing in the denominator), using the foundational rule:

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

Example: Evaluate $\lim_{x \to \infty} \frac{6x^3 - 4x + 1}{2x^3 + 5x^2 - 7}$. Divide every term by $x^3$: $\lim_{x \to \infty} \frac{6 - \frac{4}{x^2} + \frac{1}{x^3}}{2 + \frac{5}{x} - \frac{7}{x^3}} = \frac{6 - 0 + 0}{2 + 0 - 0} = \frac{6}{2} = 3$

---

4. Radical Functions at Infinity: The Negative Infinity Trap

The single most common trap in CLEP limit problems involves square roots evaluated as $x \to -\infty$. Recall the fundamental algebraic identity:

$\sqrt{x^2} = |x| = \begin{cases} x & \text{if } x > 0 \\ -x & \text{if } x < 0 \end{cases}$

When $x \to -\infty$, $x$ is negative, so $\mathbf{x = -\sqrt{x^2}}$ and $\mathbf{\sqrt{x^2} = -x}$.

Step-by-Step Worked Example: Two Distinct Horizontal Asymptotes

Problem: Find all horizontal asymptotes of $f(x) = \frac{\sqrt{16x^2 + 5}}{2x - 3}$.

Step 1: Evaluate limit as $x \to +\infty$. For $x > 0$, $x = \sqrt{x^2}$. Divide numerator by $\sqrt{x^2}$ and denominator by $x$: $\lim_{x \to \infty} \frac{\frac{\sqrt{16x^2 + 5}}{\sqrt{x^2}}}{\frac{2x - 3}{x}} = \lim_{x \to \infty} \frac{\sqrt{16 + \frac{5}{x^2}}}{2 - \frac{3}{x}} = \frac{\sqrt{16 + 0}}{2 - 0} = \frac{4}{2} = 2$

Step 2: Evaluate limit as $x \to -\infty$. For $x < 0$, $x = -\sqrt{x^2} \implies \sqrt{x^2} = -x$. Dividing the numerator by $\sqrt{x^2}$ and denominator by $x$ introduces a negative sign: $\lim_{x \to -\infty} \frac{\sqrt{16 + \frac{5}{x^2}}}{-\left(2 - \frac{3}{x}\right)} = \frac{\sqrt{16 + 0}}{-(2 - 0)} = \frac{4}{-2} = -2$

Conclusion: The function possesses two distinct horizontal asymptotes: $y = 2$ and $y = -2$.

---

5. Transcendental Functions at Infinity

Exponential Functions

• $\lim_{x \to \infty} e^x = \infty$ and $\lim_{x \to -\infty} e^x = 0$ (Horizontal asymptote $y = 0$ as $x \to -\infty$).
• $\lim_{x \to \infty} e^{-x} = 0$ and $\lim_{x \to -\infty} e^{-x} = \infty$.

Example: Evaluate $\lim_{x \to \infty} \frac{5e^x + 3}{2e^x - 7}$ vs. $\lim_{x \to -\infty} \frac{5e^x + 3}{2e^x - 7}$.

• As $x \to \infty$: Divide by $e^x \implies \lim_{x \to \infty} \frac{5 + 3e^{-x}}{2 - 7e^{-x}} = \frac{5+0}{2-0} = \frac{5}{2}$.
• As $x \to -\infty$: $e^x \to 0 \implies \frac{5(0) + 3}{2(0) - 7} = -\frac{3}{7}$.

Logarithmic & Inverse Trig Functions

• Natural Logarithm: $\lim_{x \to \infty} \ln(x) = \infty$ and $\lim_{x \to 0^+} \ln(x) = -\infty$ (Vertical asymptote at $x = 0$).
• Inverse Tangent: $\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$ (Two horizontal asymptotes: $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$).

---

6. Slant (Oblique) Asymptotes

A rational function $f(x) = \frac{P(x)}{Q(x)}$ has a slant (oblique) asymptote if and only if:

$\operatorname{deg}(P) = \operatorname{deg}(Q) + 1$

(The numerator degree is exactly one greater than the denominator degree.)

Finding the Slant Asymptote via Polynomial Long Division

Perform polynomial division to express $f(x)$ as a linear quotient plus a proper rational remainder:

$f(x) = (mx + b) + \frac{R(x)}{Q(x)}$

Since $\operatorname{deg}(R) < \operatorname{deg}(Q)$, $\lim_{x \to \pm\infty} \frac{R(x)}{Q(x)} = 0$. Therefore, as $x \to \pm\infty$, the curve approaches the line $\mathbf{y = mx + b}$.

Worked Example: Slant Asymptote

Problem: Find the slant asymptote of $f(x) = \frac{2x^3 - 3x^2 + 4}{x^2 + 1}$.

Divide $(2x^3 - 3x^2 + 4)$ by $(x^2 + 1)$:

1. $2x^3 / x^2 = 2x$. Multiply: $2x(x^2 + 1) = 2x^3 + 2x$. Subtract: $(-3x^2 - 2x + 4)$.
2. $-3x^2 / x^2 = -3$. Multiply: $-3(x^2 + 1) = -3x^2 - 3$. Subtract: $(-2x + 7)$.

Rewrite: $f(x) = (2x - 3) + \frac{-2x + 7}{x^2 + 1}$

Since $\lim_{x \to \pm\infty} \frac{-2x + 7}{x^2 + 1} = 0$, the slant asymptote is $\mathbf{y = 2x - 3}$.