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The task asks to determine whether the following limits exist (including improper limits), and to calculate them without using L'Hôpital's rule:

### Given:
\[
\lim_{x \to 0^+} \sqrt{x^2 - x}, \quad \lim_{x \to 0^-} \sqrt{x^2 - x}
\]

We need to calculate the right-hand limit (as \(x \to 0^+\)) and the left-hand limit (as \(x \to 0^-\)) for the expression \(\sqrt{x^2 - x}\).

### 1. **Right-hand limit** \(\lim_{x \to 0^+} \sqrt{x^2 - x}\):

- For \(x > 0\), the expression inside the square root is \(x^2 - x = x(x - 1)\).
- As \(x \to 0^+\), the term \(x(x - 1)\) approaches \(0\), but stays negative since \(x - 1 < 0\). Thus, for small positive \(x\), the term \(x(x - 1)\) is negative, making the expression inside the square root negative.
- Since the square root of a negative number is not real, the limit does not exist in the real number system as \(x \to 0^+\).

### 2. **Left-hand limit** \(\lim_{x \to 0^-} \sqrt{x^2 - x}\):

- For \(x < 0\), the expression \(x^2 - x = x(x - 1)\).
- As \(x \to 0^-\), \(x(x - 1)\) again approaches \(0\), but now it's positive because \(x\) is negative and \(x - 1 < 0\), making the product positive.
- For small negative \(x\), the term inside the square root is positive, and the square root of a positive number exists in the real number system.
- Thus, we evaluate the limit as \(x \to 0^-:\)
  \[
  \lim_{x \to 0^-} \sqrt{x^2 - x} = \sqrt{0} = 0.
  \]

### Conclusion:
- The right-hand limit \(\lim_{x \to 0^+} \sqrt{x^2 - x}\) **does not exist** in the real numbers.
- The left-hand limit \(\lim_{x \to 0^-} \sqrt{x^2 - x} = 0\).

Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:
1. How would this limit behave if we allowed complex numbers?
2. Can L'Hôpital's rule help in evaluating limits that result in indeterminate forms?
3. What happens to the expression if you replace \(x^2 - x\) with a different quadratic expression?
4. How can we identify if a limit does not exist in the real number system?
5. Could we approach this problem using a series expansion near \(x = 0\)?

**Tip:** Always check whether the expression inside a square root is non-negative before evaluating limits.

Avgör huruvida följande (även oegentliga) gränsvärden existerar och om så, bestäm dem (utan hjälp av L'Hopitals regler): lim sqrt(x^2 - x), x -> 0⁺, lim sqrt(x^2 - x), x -> 0⁻

This problem evaluates the right-hand and left-hand limits of the expression sqrt(x^2 - x) as x approaches 0, without using L'Hopital's rule. Explore concepts of limits, square roots, and continuity.

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