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 $\lim<em>{x \to \infty} f(x) = L$ or $\lim</em>{x \to -\infty} f(x) = L$.

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:

   $\frac{\lim<em>{x \to \infty} 3 + 5 \lim</em>{x \to \infty} \frac{1}{x} + \lim<em>{x \to \infty} \frac{1}{x^2} - 2 \lim</em>{x \to \infty} \frac{1}{x^3} + \lim<em>{x \to \infty} \frac{1}{x^4}}{\lim</em>{x \to \infty} 6 - 3 \lim<em>{x \to \infty} \frac{1}{x^2} + 2 \lim</em>{x \to \infty} \frac{1}{x^4}}$

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:

   $\lim<em>{x \to \infty} \frac{(9x^2 + x) - (3x)^2}{\sqrt{9x^2 + x} + 3x} = \lim</em>{x \to \infty} \frac{9x^2 + x - 9x^2}{\sqrt{9x^2 + x} + 3x} = \lim_{x \to \infty} \frac{x}{\sqrt{9x^2 + x} + 3x}$

3. Divide the numerator and denominator by $x$:

   $\lim<em>{x \to \infty} \frac{\frac{x}{x}}{\frac{\sqrt{9x^2 + x}}{x} + \frac{3x}{x}} = \lim</em>{x \to \infty} \frac{1}{\sqrt{\frac{9x^2 + x}{x^2}} + 3} = \lim_{x \to \infty} \frac{1}{\sqrt{9 + \frac{1}{x}} + 3}$

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