Limits at Infinity, Horizontal Asymptotes, and Vertical Asymptotes

End Behavior and Infinite Limits

  • Definition of End Behavior:

    • Analyzing limits where xx grows unbounded evaluates the end behavior of a function.

    • End behavior describes the ultimate long-term trajectory of a function as input values xx approach positive infinity (x→∞x \rightarrow \infty) or negative infinity (x→−∞x \rightarrow -\infty).

  • Unbounded Growth Examples:

    • Example 1:

    • Problem: Evaluate lim⁡x→∞2x2\lim_{x \rightarrow \infty} 2x^2

    • Analysis: As xx increases without bound towards ∞\infty, x2x^2 grows infinitely large. Multiplying by 22 maintains this unbounded positive growth.

    • Result: lim⁡x→∞2x2=∞\lim_{x \rightarrow \infty} 2x^2 = \infty (The limit does not exist as a finite real number).

    • Example 2:

    • Problem: Evaluate lim⁡x→∞1x\lim_{x \rightarrow \infty} \frac{1}{x}

    • Analysis: As xx approaches ∞\infty, the fraction 1x\frac{1}{x} has a fixed numerator and an infinitely growing denominator, driving the quotient arbitrarily close to 00.

    • Result: lim⁡x→∞1x=0\lim_{x \rightarrow \infty} \frac{1}{x} = 0

    • Concept: The finite real number 00 that the function approaches is the value of the limit, which defines a horizontal line that the graph approaches at extreme values of xx.

Limits at Infinity and Horizontal Asymptotes

Limits at Infinity and Horizontal Asymptote Definitions
  • Formal Theorems for Limits at Infinity:

    • For any positive real number nn:     lim⁡x→∞1xn=0\lim_{x \rightarrow \infty} \frac{1}{x^n} = 0     lim⁡x→−∞1xn=0\lim_{x \rightarrow -\infty} \frac{1}{x^n} = 0

  • Definition of Horizontal Asymptote:

    • The line Y=LY = L (or y=Ly = L) is defined as a horizontal asymptote of the function y=f(x)y = f(x) if either of the following statement conditions holds true:     lim⁡x→∞f(x)=L\lim_{x \rightarrow \infty} f(x) = L     lim⁡x→−∞f(x)=L\lim_{x \rightarrow -\infty} f(x) = L

  • Worked Examples for Rational and Exponential Functions:

    • Example 3:

    • Problem: Find lim⁡x→∞8x+63x−1\lim_{x \rightarrow \infty} \frac{8x + 6}{3x - 1}

    • Procedure: Divide every term in the numerator and denominator by the highest power of xx present in the denominator (which is x1x^1):       lim⁡x→∞8xx+6x3xx−1x=lim⁡x→∞8+6x3−1x\lim_{x \rightarrow \infty} \frac{\frac{8x}{x} + \frac{6}{x}}{\frac{3x}{x} - \frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{8 + \frac{6}{x}}{3 - \frac{1}{x}}

    • Evaluation: Applying the rule lim⁡x→∞1x=0\lim_{x \rightarrow \infty} \frac{1}{x} = 0 gives:       8+03−0=83\frac{8 + 0}{3 - 0} = \frac{8}{3}

    • Conclusion: The limit is 83\frac{8}{3}, and the function has a horizontal asymptote at y=83y = \frac{8}{3}.

    • Example 4:

    • Problem: Find lim⁡x→∞8x+63x2−1\lim_{x \rightarrow \infty} \frac{8x + 6}{3x^2 - 1}

    • Procedure: Divide all terms by the highest power of xx in the denominator (x2x^2):       lim⁡x→∞8xx2+6x23x2x2−1x2=lim⁡x→∞8x+6x23−1x2\lim_{x \rightarrow \infty} \frac{\frac{8x}{x^2} + \frac{6}{x^2}}{\frac{3x^2}{x^2} - \frac{1}{x^2}} = \lim_{x \rightarrow \infty} \frac{\frac{8}{x} + \frac{6}{x^2}}{3 - \frac{1}{x^2}}

    • Evaluation: Substituting limits for each reciprocal power:       0+03−0=03=0\frac{0 + 0}{3 - 0} = \frac{0}{3} = 0

    • Conclusion: The limit is 00, corresponding to a horizontal asymptote at y=0y = 0.

    • Example 5 (Biological Application - Female Arctic Foxes):

    • Context: The age-weight relationship of female Arctic Foxes caught in Svalbard, Norway, can be estimated by the function:       M(t)=3102e−e−0.022(t−56)M(t) = 3102 e^{-e^{-0.022(t - 56)}}       where tt represents the age of the foxes in days, and M(t)M(t) represents the weight of the foxes in grams.

    • Objective: Use M(t)M(t) to estimate the largest size (weight) that a female fox can attain as age tt grows indefinitely (t→∞t \rightarrow \infty).

    • Procedure:

      • Evaluate the inner exponent as t→∞t \rightarrow \infty:         −0.022(t−56)→−∞-0.022(t - 56) \rightarrow -\infty

      • Evaluate the secondary exponential term:         e−0.022(t−56)→0e^{-0.022(t - 56)} \rightarrow 0

      • Substitute back into the outer exponential structure:         −e−0.022(t−56)→−0=0-e^{-0.022(t - 56)} \rightarrow -0 = 0

      • Compute the outer exponential function value:         e0=1e^0 = 1

      • Compute the final limit of weight M(t)M(t):         lim⁡t→∞M(t)=3102×1=3102 grams\lim_{t \rightarrow \infty} M(t) = 3102 \times 1 = 3102\text{ grams}

    • Conclusion: The maximum achievable weight (largest size) for a female Arctic Fox under this model is 3102 grams3102\text{ grams}.

Shortcut Rules for Calculating Limits of Rational Functions as x Approaches Infinity

  • Rational Function Structure:

    • For a rational function defined as f(x)=N(x)D(x)f(x) = \frac{N(x)}{D(x)}, where N(x)N(x) is the numerator polynomial and D(x)D(x) is the denominator polynomial:

  • Rule 1: Higher Degree in Denominator:

    • When the degree of the denominator D(x)D(x) is strictly larger than the degree of the numerator N(x)N(x):     lim⁡x→∞N(x)D(x)=0\lim_{x \rightarrow \infty} \frac{N(x)}{D(x)} = 0

  • Rule 2: Higher Degree in Numerator:

    • When the degree of the numerator N(x)N(x) is strictly larger than the degree of the denominator D(x)D(x):     lim⁡x→∞N(x)D(x)=∞ or −∞\lim_{x \rightarrow \infty} \frac{N(x)}{D(x)} = \infty \text{ or } -\infty

  • Rule 3: Equal Degrees:

    • When the degree of the denominator D(x)D(x) is equal to the degree of the numerator N(x)N(x):     lim⁡x→∞N(x)D(x)=the quotient of the leading coefficients\lim_{x \rightarrow \infty} \frac{N(x)}{D(x)} = \text{the quotient of the leading coefficients}

Vertical Asymptotes and Infinite Limits at Finite Points

  • Evaluation of Infinite Limits at a Point:

    • Example 6:

    • Problem: Find lim⁡x→0120x2\lim_{x \rightarrow 0} \frac{1}{20x^2}

    • Procedure and Analysis:

      • Evaluate the behavior of the expression as xx approaches 00 from both sides (x→0+x \rightarrow 0^+ and x→0−x \rightarrow 0^-).

      • Since x2>0x^2 > 0 for all x≠0x \neq 0, the denominator term 20x220x^2 remains strictly positive and approaches 00

      • Taking the reciprocal of positive values approaching zero yields arbitrarily large positive values:         lim⁡x→0+120x2=∞\lim_{x \rightarrow 0^+} \frac{1}{20x^2} = \infty         lim⁡x→0−120x2=∞\lim_{x \rightarrow 0^-} \frac{1}{20x^2} = \infty

    • Result: lim⁡x→0120x2=∞\lim_{x \rightarrow 0} \frac{1}{20x^2} = \infty (Does Not Exist, DNE).

    • Graphical Behavior at x=0x = 0: At x=0x = 0, the graph shoots vertically upwards towards positive infinity (∞\infty) from both the left and right sides.

Vertical Asymptote Definitions
  • Formal Definition of Vertical Asymptote:

    • The vertical line x=ax = a is called a vertical asymptote of the function y=f(x)y = f(x) if at least one of the following conditions is true:

    • lim⁡x→af(x)=∞,DNE\lim_{x \rightarrow a} f(x) = \infty, \text{DNE}

    • lim⁡x→af(x)=−∞,DNE\lim_{x \rightarrow a} f(x) = -\infty, \text{DNE}

    • lim⁡x→a−f(x)=∞,DNE\lim_{x \rightarrow a^-} f(x) = \infty, \text{DNE}

    • lim⁡x→a−f(x)=−∞,DNE\lim_{x \rightarrow a^-} f(x) = -\infty, \text{DNE}

    • lim⁡x→a+f(x)=∞,DNE\lim_{x \rightarrow a^+} f(x) = \infty, \text{DNE}

    • lim⁡x→a+f(x)=−∞,DNE\lim_{x \rightarrow a^+} f(x) = -\infty, \text{DNE}

  • Comprehensive Asymptote Identification:

    • Example 7:

    • Problem: Identify all asymptotes of the function f(x)=2x−3x+1f(x) = \frac{2x - 3}{x + 1}

    • Vertical Asymptote Analysis:

      • Set the denominator equal to zero to find candidate vertical discontinuities:         x+1=0  ⟹  x=−1x + 1 = 0 \implies x = -1

      • Test one-sided limits as x→−1x \rightarrow -1:

      • As x→−1+x \rightarrow -1^+, numerator approaches 2(−1)−3=−52(-1) - 3 = -5, and denominator approaches 0+0^+ (small positive numbers), giving:           lim⁡x→−1+2x−3x+1=−∞\lim_{x \rightarrow -1^+} \frac{2x - 3}{x + 1} = -\infty

      • As x→−1−x \rightarrow -1^-, numerator approaches −5-5, and denominator approaches 0−0^- (small negative numbers), giving:           lim⁡x→−1−2x−3x+1=∞\lim_{x \rightarrow -1^-} \frac{2x - 3}{x + 1} = \infty

      • Conclusion for Vertical Asymptote: The vertical line x=−1x = -1 is a vertical asymptote.

    • Horizontal Asymptote Analysis:

      • Calculate limits at infinity:         lim⁡x→∞2x−3x+1=lim⁡x→∞2−3x1+1x=2−01+0=2\lim_{x \rightarrow \infty} \frac{2x - 3}{x + 1} = \lim_{x \rightarrow \infty} \frac{2 - \frac{3}{x}}{1 + \frac{1}{x}} = \frac{2 - 0}{1 + 0} = 2         lim⁡x→−∞2x−3x+1=lim⁡x→−∞2−3x1+1x=2−01+0=2\lim_{x \rightarrow -\infty} \frac{2x - 3}{x + 1} = \lim_{x \rightarrow -\infty} \frac{2 - \frac{3}{x}}{1 + \frac{1}{x}} = \frac{2 - 0}{1 + 0} = 2

      • Conclusion for Horizontal Asymptote: The line y=2y = 2 is a horizontal asymptote.