Limit X 2 0Ec2A0

1. **State the problem:** We need to find the limit of the function $$\frac{x-3}{x-2}$$ as $x$ approaches $2$. 2. **Recall the limit concept:** The limit of a function as $x$ approaches a value $a$ is the value that the function approaches as $x$ gets closer to $a$. 3. **Check direct substitution:** Substitute $x=2$ into the function: $$\frac{2-3}{2-2} = \frac{-1}{0}$$ This is undefined because division by zero is not allowed. 4. **Analyze the behavior near $x=2$:** - For $x$ slightly less than $2$, say $x=1.9$, denominator $x-2$ is negative and close to zero, numerator $x-3$ is negative. - For $x$ slightly greater than $2$, say $x=2.1$, denominator $x-2$ is positive and close to zero, numerator $x-3$ is negative. 5. **Determine one-sided limits:** - Left-hand limit as $x \to 2^-$: $$\lim_{x \to 2^-} \frac{x-3}{x-2} = \frac{\text{negative}}{\text{small negative}} = \text{positive large} = +\infty$$ - Right-hand limit as $x \to 2^+$: $$\lim_{x \to 2^+} \frac{x-3}{x-2} = \frac{\text{negative}}{\text{small positive}} = \text{negative large} = -\infty$$ 6. **Conclusion:** Since the left-hand and right-hand limits are not equal, the limit does not exist at $x=2$. **Final answer:** $$\lim_{x \to 2} \frac{x-3}{x-2} \text{ does not exist.}$$