$$ \lim_{x\to\infty}\frac{\sqrt{x^2-3x}}{x-3} $$

Let's analyze the given limit:
\[
\lim_{x\to\infty}\frac{\sqrt{x^2-3x}}{x-3}.
\]

### Step 1: Simplify the expression inside the square root
We can factor out $x^2$ from the expression inside the square root:
\[
\sqrt{x^2 - 3x} = \sqrt{x^2\left(1 - \frac{3}{x}\right)} = x\sqrt{1 - \frac{3}{x}}.
\]

### Step 2: Rewrite the original expression using the simplification
Substituting this into the limit, we have:
\[
\lim_{x\to\infty} \frac{x\sqrt{1 - \frac{3}{x}}}{x - 3}.
\]

### Step 3: Factor $x$ out of the denominator
The denominator can be rewritten as:
\[
x - 3 = x\left(1 - \frac{3}{x}\right).
\]

Now, substituting both simplified expressions, the limit becomes:
\[
\lim_{x\to\infty} \frac{x\sqrt{1 - \frac{3}{x}}}{x\left(1 - \frac{3}{x}\right)}.
\]

### Step 4: Simplify the fraction
We can cancel the $x$ terms in the numerator and denominator:
\[
\lim_{x\to\infty} \frac{\sqrt{1 - \frac{3}{x}}}{1 - \frac{3}{x}}.
\]

### Step 5: Evaluate the limit as $x$ approaches infinity
As $x$ approaches infinity, the terms involving $\frac{3}{x}$ approach zero:
\[
\lim_{x\to\infty} \frac{\sqrt{1 - 0}}{1 - 0} = \frac{1}{1} = 1.
\]

### Conclusion
The value of the limit is:
\[
\lim_{x\to\infty}\frac{\sqrt{x^2-3x}}{x-3} = 1.
\]

Would you like further details on this solution, or do you have any other questions?

Here are some related questions to deepen your understanding:
1. What happens if the expression inside the square root is slightly modified?
2. How does the limit change if the denominator is $x + 3$ instead of $x - 3$?
3. How can L'Hôpital's Rule be used to solve similar limits involving indeterminate forms?
4. What are other common techniques for evaluating limits involving square roots?
5. How would you approach the limit if it were $x$ approaching negative infinity?

**Tip:** When dealing with limits involving square roots, it is often helpful to factor out the highest power of the variable to simplify the expression.

Explore the step-by-step process to evaluate the limit of \frac{\sqrt{x^2 - 3x}}{x - 3} as x approaches infinity. This solution involves algebraic simplifications, limit laws, and factorization techniques, perfect for students in Grades 11-12 or early college calculus.

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