Limits at Infinity and Horizontal Asymptotes

Limits at Infinity and Horizontal Asymptotes

Definition of Limit at Infinity

  • Formal Definition: If f is a function defined on some interval \left[ a, \infty \right), then \lim_{x \to \infty} f(x) = L means that the values of f(x) can be made arbitrarily close to L by requiring x to be sufficiently large.

Example: Estimating \lim_{x \to \infty} \frac{1}{x}

  • Consider the function f(x) = \frac{1}{x}.

  • We want to estimate \lim_{x \to \infty} \frac{1}{x}.

  • Create a table of values for large x:

    x

    f(x) = \frac{1}{x}

    10

    0.1

    100

    0.01

    1,000

    0.001

    10,000

    0.0001

    100,000

    0.00001

    1,000,000

    0.000001

  • As x approaches infinity, \frac{1}{x} approaches 0.

  • Therefore, \lim_{x \to \infty} \frac{1}{x} = 0.

Definition of Limit at Negative Infinity

  • If f is a function defined on some interval \left( -\infty, a \right], then \lim_{x \to -\infty} f(x) = L means that the values of f(x) can be made arbitrarily close to L by requiring x to be sufficiently large negative.

Example: Estimating \lim_{x \to -\infty} \frac{x+1}{2x-3}

  • Consider the function f(x) = \frac{x+1}{2x-3}.

  • We want to estimate \lim_{x \to -\infty} \frac{x+1}{2x-3}.

  • Create a table of values for large negative x:

    x

    f(x) = \frac{x+1}{2x-3}

    Approximation

    -10

    \frac{-9}{-23} = \frac{9}{23}

    0.391

    -100

    \frac{-99}{-203} = \frac{99}{203}

    0.488

    -1,000

    \frac{-999}{-2003} = \frac{999}{2003}

    0.499

    -10,000

    \frac{-9999}{-20003} = \frac{9999}{20003}

    0.49995

  • As x approaches negative infinity, \frac{x+1}{2x-3} approaches 0.5.

  • Therefore, \lim_{x \to -\infty} \frac{x+1}{2x-3} = 0.5.

Definition of Horizontal Asymptote

  • The line y = L is called a horizontal asymptote of the curve y = f(x) if either \lim{x \to \infty} f(x) = L or \lim{x \to -\infty} f(x) = L.

Examples of Horizontal Asymptotes

  1. The line y = 0 is a horizontal asymptote of the curve f(x) = \frac{1}{x} because \lim_{x \to \infty} \frac{1}{x} = 0.

  2. The line y = \frac{1}{2} is a horizontal asymptote of the curve f(x) = \frac{x+1}{2x-3} because \lim_{x \to -\infty} \frac{x+1}{2x-3} = \frac{1}{2}.

Theorem

  • If r > 0 is a rational number, then \lim_{x \to \infty} \frac{1}{x^r} = 0.

  • If r > 0 is a rational number such that x^r is defined for all x, then \lim_{x \to -\infty} \frac{1}{x^r} = 0.

Examples Illustrating the Theorem

  1. \lim_{x \to \infty} \frac{1}{x} = 0 because r = 1 is a positive rational number.

  2. \lim_{x \to -\infty} \frac{1}{x} = 0 because r = 1 is a positive rational number and x^1 = x is defined for all x.

  3. \lim_{x \to \infty} \frac{1}{x^{0.5}} = 0 because r = 0.5 is a positive rational number.

  4. \lim_{x \to -\infty} \frac{1}{x^{0.5}} does not exist because x^{0.5} = \sqrt{x} is not defined for x < 0.

Example: Evaluating \lim_{x \to \infty} \frac{3x^4 + 5x^3 + x^2 - 2x + 1}{6x^4 - 3x^2 + 2}

  1. Divide each term in the numerator and denominator by x^4 (the highest power of x in the denominator):

    \lim_{x \to \infty} \frac{\frac{3x^4}{x^4} + \frac{5x^3}{x^4} + \frac{x^2}{x^4} - \frac{2x}{x^4} + \frac{1}{x^4}}{\frac{6x^4}{x^4} - \frac{3x^2}{x^4} + \frac{2}{x^4}}

  2. Simplify:

    \lim_{x \to \infty} \frac{3 + \frac{5}{x} + \frac{1}{x^2} - \frac{2}{x^3} + \frac{1}{x^4}}{6 - \frac{3}{x^2} + \frac{2}{x^4}}

  3. Apply the limit to each term:

    \frac{\lim{x \to \infty} 3 + 5 \lim{x \to \infty} \frac{1}{x} + \lim{x \to \infty} \frac{1}{x^2} - 2 \lim{x \to \infty} \frac{1}{x^3} + \lim{x \to \infty} \frac{1}{x^4}}{\lim{x \to \infty} 6 - 3 \lim{x \to \infty} \frac{1}{x^2} + 2 \lim{x \to \infty} \frac{1}{x^4}}

  4. Evaluate the limits (using the theorem):

    \frac{3 + 5(0) + 0 - 2(0) + 0}{6 - 3(0) + 2(0)} = \frac{3}{6} = \frac{1}{2}

  • Therefore, \lim_{x \to \infty} \frac{3x^4 + 5x^3 + x^2 - 2x + 1}{6x^4 - 3x^2 + 2} = \frac{1}{2}.

Example: Evaluating \lim_{x \to \infty} (\sqrt{9x^2 + x} - 3x)

  1. Multiply by the conjugate:

    \lim_{x \to \infty} (\sqrt{9x^2 + x} - 3x) \cdot \frac{\sqrt{9x^2 + x} + 3x}{\sqrt{9x^2 + x} + 3x}

  2. Simplify:

    \lim{x \to \infty} \frac{(9x^2 + x) - (3x)^2}{\sqrt{9x^2 + x} + 3x} = \lim{x \to \infty} \frac{9x^2 + x - 9x^2}{\sqrt{9x^2 + x} + 3x} = \lim_{x \to \infty} \frac{x}{\sqrt{9x^2 + x} + 3x}

  3. Divide the numerator and denominator by x:

    \lim{x \to \infty} \frac{\frac{x}{x}}{\frac{\sqrt{9x^2 + x}}{x} + \frac{3x}{x}} = \lim{x \to \infty} \frac{1}{\sqrt{\frac{9x^2 + x}{x^2}} + 3} = \lim_{x \to \infty} \frac{1}{\sqrt{9 + \frac{1}{x}} + 3}

  4. Evaluate the limit:

    \frac{1}{\sqrt{9 + \lim_{x \to \infty} \frac{1}{x}} + 3} = \frac{1}{\sqrt{9 + 0} + 3} = \frac{1}{3 + 3} = \frac{1}{6}

  • Therefore, \lim_{x \to \infty} (\sqrt{9x^2 + x} - 3x) = \frac{1}{6}.