MA 161 - Lesson 6: Infinite Limits & Vertical Asymptotes

Warm-up: Evaluating Large Numbers

Consider the following calculations to understand the concept of infinity:

\(\frac{1}{0.1} = 10\)
\(\frac{1}{0.01} = 100\)
\(\frac{1}{0.0001} = 10,000\)
\(\frac{1}{-0.1} = -10\)
\(\frac{1}{-0.01} = -100\)
\(\frac{1}{-0.0001} = -10,000\)

Key Idea: As the denominator gets smaller and smaller, the fraction gets larger and larger, approaching infinity.

Infinite Limits

Consider the function \(f(x) = \frac{1}{x-7}\) as \(x\) approaches 7.

Right-Hand Limit

As \(x\) approaches 7 from the right (\(x \to 7^+\)):

Let's examine what happens as \(x\) gets closer to 7 from values greater than 7:

When \(x = 7.01\): \(f(x) = \frac{1}{7.01-7} = \frac{1}{0.01} = 100\)
When \(x = 7.001\): \(f(x) = \frac{1}{0.001} = 1000\)
When \(x = 7.0001\): \(f(x) = \frac{1}{0.0001} = 10,000\)

Since \(x - 7 > 0\) when approaching from the right, we get larger and larger positive values:

\[\lim_{x \to 7^+} f(x) = +\infty\]

Left-Hand Limit

As \(x\) approaches 7 from the left (\(x \to 7^-\)):

Let's examine what happens as \(x\) gets closer to 7 from values less than 7:

When \(x = 6.99\): \(f(x) = \frac{1}{6.99-7} = \frac{1}{-0.01} = -100\)
When \(x = 6.999\): \(f(x) = \frac{1}{-0.001} = -1000\)
When \(x = 6.9999\): \(f(x) = \frac{1}{-0.0001} = -10,000\)

Since \(x - 7 < 0\) when approaching from the left, we get larger and larger negative values:

\[\lim_{x \to 7^-} f(x) = -\infty\]

Graph Description: The graph of \(f(x) = \frac{1}{x-7}\) near \(x = 7\) shows a vertical asymptote at \(x = 7\). The function shoots up to positive infinity on the right side of the line and down to negative infinity on the left side.

Since the right and left limits are different, \(\lim_{x \to 7} f(x) = \text{DNE}\) (does not exist).

Vertical Asymptotes

Definition: For any function \(f(x)\), the line \(x = a\) is a vertical asymptote if:

\[\lim_{x \to a^+} f(x) = \pm\infty \quad \text{OR} \quad \lim_{x \to a^-} f(x) = \pm\infty\]

Therefore, \(x = 7\) is a vertical asymptote for \(f(x) = \frac{1}{x-7}\).

Examples of Finding Vertical Asymptotes

Example 1: Rational Function with Non-Zero Numerator

Find vertical asymptotes for \(f(x) = \frac{x^2 - 4x - 28}{x - 7}\)

Solution:

First, check what happens when we substitute \(x = 7\):

Numerator: \(7^2 - 4(7) - 28 = 49 - 28 - 28 = -7 \neq 0\)
Denominator: \(7 - 7 = 0\)

Since the numerator is non-zero and the denominator is zero, we have a vertical asymptote.

Analyzing the behavior:

As \(x \to 7^+\): \(\frac{-7}{\text{small positive}} = -\infty\)
As \(x \to 7^-\): \(\frac{-7}{\text{small negative}} = +\infty\)

Therefore, \(x = 7\) is a vertical asymptote.

Example 2: Quadratic Denominator

Does \(f(x) = \frac{1}{(x-7)^2}\) have a vertical asymptote?

Solution:

Since \((x-7)^2\) is always positive (except when \(x = 7\)), regardless of direction of approach:

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

Therefore, \(x = 7\) is a vertical asymptote.

Example 3: When Factors Cancel

Evaluate \(\lim_{x \to 1} \frac{x^2 - 4x + 3}{x - 1}\)

Solution:

Factor the numerator: \(x^2 - 4x + 3 = (x-3)(x-1)\)

So: \(f(x) = \frac{(x-3)(x-1)}{x-1} = x-3\) (for \(x \neq 1\))

Since both numerator and denominator equal 0 when \(x = 1\), we can cancel the common factor.

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

Warning: \(x = 1\) is NOT a vertical asymptote because the limit exists and equals -2. Be careful when both numerator and denominator equal zero.

General Problem: Finding Parameter Values

For what values of \(a\) does \(f(x) = \frac{x^2 - 3x - 28}{(x-a)(x-1)}\) have a vertical asymptote at \(x = a\)?

Solution:

First, factor the numerator: \(x^2 - 3x - 28 = (x-7)(x+4)\)

So: \(f(x) = \frac{(x-7)(x+4)}{(x-a)(x-1)}\)

Case Analysis:

If \(a = -4\):
\(f(x) = \frac{(x-7)(x+4)}{(x+4)(x-1)} = \frac{x-7}{x-1}\) (for \(x \neq -4\))

At \(x = -4\): \(\lim_{x \to -4} f(x) = \frac{-4-7}{-4-1} = \frac{-11}{-5} = \frac{11}{5}\)

\(x = -4\) is NOT a vertical asymptote.
If \(a = 7\):
\(f(x) = \frac{(x-7)(x+4)}{(x-7)(x-1)} = \frac{x+4}{x-1}\) (for \(x \neq 7\))

At \(x = 7\): \(\lim_{x \to 7} f(x) = \frac{7+4}{7-1} = \frac{11}{6}\)

\(x = 7\) is NOT a vertical asymptote.
For any \(a \neq 7\) and \(a \neq -4\):
There are no common factors to cancel, so there is a vertical asymptote at \(x = a\).

Conclusion: The function \(f(x)\) has a vertical asymptote at \(x = a\) for any value \(a \neq 7\) and \(a \neq -4\).