Given $g(x) = \frac{x-2}{(x-1)^2(x-3)}$, which of the following statements about the limit of g(x) as x approaches 1 is correct? A) $\lim_{x \to 1} g(x) = 0$ B) The limit does not... Given $g(x) = \frac{x-2}{(x-1)^2(x-3)}$, which of the following statements about the limit of g(x) as x approaches 1 is correct? A) $\lim_{x \to 1} g(x) = 0$ B) The limit does not exist C) $\lim_{x \to 1} g(x) = \infty$ D) $\lim_{x \to 1} g(x) = -\infty$
Understand the Problem
The question asks us to evaluate the limit of the function g(x) as x approaches 1. We need to determine whether the limit exists and, if it does, find its value. Since direct substitution results in a zero in the denominator, we need to analyze the behavior of the function as x approaches 1 from the left and right.
Answer
The limit does not exist.
Answer for screen readers
The limit does not exist.
Steps to Solve
- Analyze the limit as x approaches 1 from the left
We need to evaluate $\lim_{x \to 1^-} g(x)$ where $g(x) = \frac{x+1}{x-1}$. As $x$ approaches 1 from the left, $x$ is slightly less than 1, so $x-1$ is a small negative number. The numerator $x+1$ approaches 2. Thus, we have a positive number divided by a small negative number, which results in negative infinity.
$$ \lim_{x \to 1^-} \frac{x+1}{x-1} = -\infty$$
- Analyze the limit as x approaches 1 from the right
We need to evaluate $\lim_{x \to 1^+} g(x)$ where $g(x) = \frac{x+1}{x-1}$. As $x$ approaches 1 from the right, $x$ is slightly greater than 1, so $x-1$ is a small positive number. The numerator $x+1$ approaches 2. Thus, we have a positive number divided by a small positive number, which results in positive infinity.
$$ \lim_{x \to 1^+} \frac{x+1}{x-1} = +\infty$$
- Determine if the limit exists
Since the limit from the left is $-\infty$ and the limit from the right is $+\infty$, the two one-sided limits are not equal. Therefore, the limit does not exist.
$$ \lim_{x \to 1} \frac{x+1}{x-1} \text{ does not exist} $$
The limit does not exist.
More Information
The function $g(x) = \frac{x+1}{x-1}$ has a vertical asymptote at $x = 1$. This means that as $x$ gets closer to 1, the function's value increases or decreases without bound.
Tips
A common mistake is to assume that because the numerator approaches 2 and the denominator approaches 0, the limit is simply undefined or 0. It's crucial to analyze the sign of the denominator as $x$ approaches 1 from both sides to determine whether the limit approaches positive or negative infinity. Another error might be trying to apply L'Hopital's rule, which is not applicable here because the limit is not in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
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