The problem is to find the limit of the function $\frac{1}{x-3}$ as $x$ approaches 3. We need to evaluate $\lim_{x \to 3} \frac{1}{x-3}$.

1. Problem Description

The problem is to find the limit of the function $\frac{1}{x-3}$ as $x$ approaches

3. We need to evaluate $\lim_{x \to 3} \frac{1}{x-3}$.

2. Solution Steps

We consider the limit as $x$ approaches 3 from the left and from the right.

First, let's consider the limit as $x$ approaches 3 from the left, denoted as $x \to 3^-$. In this case, $x$ is slightly less than 3, so $x-3$ is a small negative number.

Therefore, $\frac{1}{x-3}$ will be a large negative number.

$$\lim_{x \to 3^-} \frac{1}{x-3} = -\infty$$

Next, let's consider the limit as $x$ approaches 3 from the right, denoted as $x \to 3^+$. In this case, $x$ is slightly greater than 3, so $x-3$ is a small positive number.

Therefore, $\frac{1}{x-3}$ will be a large positive number.

$$\lim_{x \to 3^+} \frac{1}{x-3} = +\infty$$

Since the left-hand limit and the right-hand limit are not equal, the limit does not exist.

3. Final Answer

The limit does not exist. We can say it diverges.

More formally, we can say the limit is undefined.