The given limit problem is:

$\lim_{x \to 2^+} \frac{x^2 - 2x}{\lfloor x \rfloor - x}$

Key aspects:

• $\lfloor x \rfloor$ represents the floor function, which gives the greatest integer less than or equal to $x$.
• We are approaching $x = 2$ from the right ($x \to 2^+$), so $x$ is slightly greater than 2.

Analyzing the limit:

1. As $x \to 2^+$, $\lfloor x \rfloor = 2$, because the floor of any number slightly greater than 2 is still 2.

2. Let's now examine the numerator: $x^2 - 2x = x(x - 2)$ As $x \to 2$, $x^2 - 2x \to 0$.

3. The denominator: $\lfloor x \rfloor - x = 2 - x$ As $x \to 2^+$, $2 - x \to 0^-$, meaning the denominator is approaching zero from the negative side.

Thus, as $x \to 2^+$, we have a 0/0 form, which suggests that we can apply L'Hôpital's Rule to evaluate the limit.

Applying L'Hôpital's Rule:

1. Differentiate the numerator: $\frac{d}{dx}(x^2 - 2x) = 2x - 2$

2. Differentiate the denominator: $\frac{d}{dx}(2 - x) = -1$

3. Now, applying L'Hôpital's Rule: $\lim_{x \to 2^+} \frac{2x - 2}{-1} = \lim_{x \to 2^+} -(2x - 2)$ Substituting $x = 2$: $-(2(2) - 2) = -(4 - 2) = -2$

Final result:

$\lim_{x \to 2^+} \frac{x^2 - 2x}{\lfloor x \rfloor - x} = -2$

Would you like a detailed explanation of L'Hôpital's Rule or any other clarifications?

Here are some related questions to deepen your understanding:

1. How does L'Hôpital's Rule work for indeterminate forms?
2. What happens if the limit approaches from the left, $x \to 2^-$?
3. How does the floor function affect limits in general?
4. Can you think of other situations where the 0/0 indeterminate form occurs?
5. What other methods besides L'Hôpital’s Rule can be used to evaluate limits?

Tip: Always check for indeterminate forms before applying L'Hôpital's Rule to ensure it is applicable.