1. The Idea of an Infinite Limit
An **infinite limit** is a limit that "does not exist" in the traditional sense, because the function's value is growing without bound towards either positive or negative infinity. This behavior occurs when the denominator of a rational function approaches zero, while the numerator approaches a non-zero number.
When a function approaches infinity or negative infinity as \( x \) approaches a specific value, this indicates the presence of a **vertical asymptote** at that value.
A function \( f(x) \) has a vertical asymptote at \( x = c \) if any of the following statements are true:
\[ \lim_{x \to c^-} f(x) = \pm \infty \] \[ \lim_{x \to c^+} f(x) = \pm \infty \]2. Finding Infinite Limits Algebraically
To find infinite limits, follow these steps:
- Start by trying to substitute the value \( c \) into the function.
- If you get a form like \( \frac{k}{0} \), where \( k \) is a non-zero number, you have an infinite limit.
- To determine whether the limit is \( +\infty \) or \( -\infty \), you must check the one-sided limits.
- Choose a test value very close to \( c \) from the left and another very close to \( c \) from the right.
- Substitute these test values into the function to determine the sign of the result.
For example, for the function \( f(x) = \frac{1}{x-2} \), the limit as \( x \to 2 \) is \( \frac{1}{0} \). This tells us the limit is infinite. To find the sign, we check the one-sided limits:
- As \( x \to 2^+ \) (e.g., \(x=2.001\)), the numerator is positive (1) and the denominator is positive (\(0.001\)). So, the limit is \( +\infty \).
- As \( x \to 2^- \) (e.g., \(x=1.999\)), the numerator is positive (1) and the denominator is negative (\(-0.001\)). So, the limit is \( -\infty \).
Since the one-sided limits are not equal, the two-sided limit \( \lim_{x \to 2} \frac{1}{x-2} \) does not exist.
3. Check Yourself
Evaluate the following limits.
Question 1: Find \( \lim_{x \to 3^+} \frac{x+1}{x-3} \).
As \( x \to 3^+ \), the numerator \( (x+1) \) approaches \( 4 \), a positive number. The denominator \( (x-3) \) approaches \( 0 \) from the positive side (e.g., a test value like \(3.001\) gives a positive result).
Therefore, the limit is a positive number divided by a small positive number, which gives:
\[ \lim_{x \to 3^+} \frac{x+1}{x-3} = +\infty \]Question 2: Find \( \lim_{x \to 0} \frac{x-1}{x^2} \).
As \( x \to 0 \), the numerator \( (x-1) \) approaches \( -1 \), a negative number. The denominator \( x^2 \) approaches \( 0 \), but since it is squared, it will always be a positive value, whether you approach from the left or the right.
Since the numerator is negative and the denominator is a small positive number, the limit is negative infinity from both sides.
\[ \lim_{x \to 0} \frac{x-1}{x^2} = -\infty \]Question 3: Find \( \lim_{x \to 5} \frac{x+2}{x^2 - 25} \).
The denominator can be factored as \( (x-5)(x+5) \).
As \( x \to 5 \), the numerator \( (x+2) \) approaches \( 7 \), which is positive.
The denominator \( (x^2 - 25) \) approaches 0. We must check the one-sided limits.
- **As \( x \to 5^+ \):** The numerator is positive. The denominator factors are \( (x-5) \to 0^+ \) and \( (x+5) \to 10 \). The result is a positive number divided by a positive number, so the limit is \( +\infty \).
- **As \( x \to 5^- \):** The numerator is positive. The denominator factors are \( (x-5) \to 0^- \) and \( (x+5) \to 10 \). The result is a positive number divided by a negative number, so the limit is \( -\infty \).
Since the one-sided limits are not equal, the two-sided limit **does not exist**.