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2.5 Limits at Infinity

(NOT on exam 1)

Let \( f(x) = \frac{1}{x} \). The limit \( \lim_{x \to a} \frac{1}{x} \) is easy if \( a \) is a number.

If \( a \) goes to a very large positive number \( \to \lim_{x \to \infty} f(x) \)
If \( a \) goes to a very large negative number \( \to \lim_{x \to -\infty} f(x) \)

\( \lim_{x \to \infty} \frac{1}{x} \to \frac{1}{\text{very large number}} \to \text{very small number} \) (practically but not exactly zero)

\( \lim_{x \to \infty} \frac{1}{x} = 0 \)

We can make \( \frac{1}{x} \) as close to 0 as we want by making \( x \) sufficiently large.

Likewise, we see

\( \lim_{x \to -\infty} \frac{1}{x} = 0 \)

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\( \lim_{x \to \pm\infty} \frac{c}{x} = \lim_{x \to \pm\infty} c \cdot \frac{1}{x} = \left( \lim_{x \to \pm\infty} c \right) \left( \lim_{x \to \pm\infty} \frac{1}{x} \right) = 0 \)

Note: \( c \) is some constant (number).

Similarly,

\( \lim_{x \to \pm\infty} \frac{c}{x^n} = \left( \lim_{x \to \pm\infty} c \right) \left( \lim_{x \to \pm\infty} \frac{1}{x^n} \right) \)

\( = \left( \lim_{x \to \pm\infty} c \right) \left( \lim_{x \to \pm\infty} \frac{1}{x} \right)^n = 0 \)

The higher the \( n \), the faster \( \frac{c}{x^n} \to 0 \).

Important Result:

\( \lim_{x \to \pm\infty} \frac{c}{x^n} = 0 \)

\( (n > 0) \)

Using this we can handle just about any rational function.

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Example: Limits at Infinity

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

First, identify the dominant term in the denominator (which is usually the term with the highest power).

Note its power.

Here, it's \( 3x^2 \), its power is 2.

Next, divide top and bottom by \( x \) to that power. In this case, we divide top and bottom by \( x^2 \).

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

\[ = \lim_{x \to \infty} \frac{4 - \left(\frac{1}{x}\right) + \left(\frac{1}{x^2}\right)}{3 + \left(\frac{5}{x}\right) - \left(\frac{3}{x^2}\right)} = \frac{4}{3} \]

As \( x \) becomes larger and larger, \( \frac{4x^2 - x + 1}{3x^2 + 5x - 3} \) becomes closer and closer to \( \frac{4}{3} \).

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We handle \( x \to -\infty \) exactly the same way:

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

For rational functions (polynomial over polynomial) the limits at \( \infty \) and \( -\infty \) are the same.

Example

\[ \lim_{x \to \infty} \frac{x - 7}{x^2 + 9} \]

Dominant term in denominator: \( x^2 \), power is 2. So we divide top and bottom by \( x^2 \).

\[ \lim_{x \to \infty} \frac{\frac{x - 7}{x^2}}{\frac{x^2 + 9}{x^2}} = \lim_{x \to \infty} \frac{\frac{x}{x^2} - \frac{7}{x^2}}{\frac{x^2}{x^2} + \frac{9}{x^2}} = \lim_{x \to \infty} \frac{\frac{1}{x} - \frac{7}{x^2}}{1 + \frac{9}{x^2}} = \frac{0}{1} = 0 \]

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Limits at Infinity: Examples and Patterns

Example

\[ \lim_{x \to \infty} \frac{x^3 + 7x^2}{x + 9} \]

Dominant term in denominator: \( x \) (power is 1). So divide top and bottom by \( x^1 \):

\[ \lim_{x \to \infty} \frac{\frac{x^3 + 7x^2}{x}}{\frac{x + 9}{x}} = \lim_{x \to \infty} \frac{x^2 + 7x}{1 + \frac{9}{x}} = \frac{\infty}{1} = \infty \]

Note: As \( x \to \infty \), the terms \( x^2 \) and \( 7x \) both approach \( \infty \), while \( \frac{9}{x} \) approaches 0.

Pattern

If numerator has greater degree, limit is \( \infty \) or \( -\infty \) as \( x \to \pm\infty \).
If denominator has greater degree, limit is 0 as \( x \to \pm\infty \).
If degrees are the same, limit as \( x \to \pm\infty \) is the ratio of the coefficients of terms w/ highest power.

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

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Graphical Interpretation

Graphically,

\( \lim_{x \to \infty} f(x) \) is the horizontal asymptote to the RIGHT.
\( \lim_{x \to -\infty} f(x) \) is the horizontal asymptote to the LEFT.

For rational functions, these will always be the same, but if root functions are involved, sometimes \( \lim_{x \to \infty} f(x) \neq \lim_{x \to -\infty} f(x) \).

Example with Absolute Value and Roots

For example, \( |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \)

\[ \sqrt{x^2} = |x| \]

So,

\[ \lim_{x \to \infty} \sqrt{x^2} = \lim_{x \to \infty} |x| = \infty \text{ because } x \to \infty \text{ means } x \geq 0 \]

\[ \lim_{x \to -\infty} \sqrt{x^2} = \lim_{x \to -\infty} |x| = -\infty \]

Note: The last line in the handwritten notes indicates a limit of \( -\infty \) for the absolute value as \( x \to -\infty \), though mathematically \( |x| \) always approaches \( \infty \).

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Example: Limit at Negative Infinity

Given the function:

\[ f(x) = \frac{3x+2}{x+\sqrt{4x^2+1}} \]

Find the limit:

\[ \lim_{x \to -\infty} f(x) = ? \]

Identifying the Dominant Term

Dominant term in denominator: \( x \) or \( \sqrt{4x^2+1} \)

As \( x \to -\infty \):

\[ \sqrt{4x^2+1} \approx \sqrt{4x^2} \approx 2|x| \]

Both terms have a power of 1. So, divide top and bottom by \( x^1 \).

Step-by-Step Calculation

\[ \lim_{x \to -\infty} \frac{3x+2}{x+\sqrt{4x^2+1}} = \lim_{x \to -\infty} \frac{\frac{3x+2}{x}}{\frac{x+\sqrt{4x^2+1}}{x}} \]

Distributing the division:

\[ = \lim_{x \to -\infty} \frac{\frac{3x}{x} + \frac{2}{x}}{\frac{x}{x} + \frac{\sqrt{4x^2+1}}{-\sqrt{x^2}}} \]

Note: We use \( -\sqrt{x^2} \) in the denominator because as \( x \to -\infty \), \( x \) is negative, and \( \sqrt{x^2} = |x| = -x \).

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Example Continued: Limit at Negative Infinity

Bottom of last page:

\[ \lim_{x \to -\infty} \frac{\frac{3x}{x} + \frac{2}{x}}{\frac{x}{x} + \frac{\sqrt{4x^2+1}}{-\sqrt{x^2}}} \]

Needs negative because \( x \to -\infty \) and \( \sqrt{x^2} = |x| \).

Simplifying the Expression

\[ = \lim_{x \to -\infty} \frac{3 + \frac{2}{x}}{1 - \sqrt{\frac{4x^2+1}{x^2}}} \]

\[ = \lim_{x \to -\infty} \frac{3 + \frac{2}{x}}{1 - \sqrt{4 + \frac{1}{x^2}}} \]

As \( x \to -\infty \), the terms \( \frac{2}{x} \to 0 \) and \( \frac{1}{x^2} \to 0 \).

Final Result

\[ = \frac{3}{1 - \sqrt{4}} = \frac{3}{1 - 2} \]\[ = -3 \]