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To solve the limit 

\[
\lim_{t \to \infty} \frac{\sqrt{t + t^2}}{3t - t^2},
\]

let's simplify the expression step-by-step:

### Step 1: Simplify the numerator
The numerator is \(\sqrt{t + t^2}\). We can factor out \(t^2\) inside the square root:

\[
\sqrt{t + t^2} = \sqrt{t^2(1 + \frac{t}{t^2})} = \sqrt{t^2(1 + \frac{1}{t})} = t\sqrt{1 + \frac{1}{t}}.
\]

As \(t \to \infty\), \(\frac{1}{t} \to 0\), so \(\sqrt{1 + \frac{1}{t}} \to 1\). Thus, the numerator simplifies to approximately \(t\).

### Step 2: Simplify the denominator
The denominator is \(3t - t^2\). Factoring out \(t^2\) from the terms inside the expression:

\[
3t - t^2 = -t^2(1 - \frac{3}{t}).
\]

As \(t \to \infty\), \(\frac{3}{t} \to 0\), so the denominator simplifies to approximately \(-t^2\).

### Step 3: Combine the simplified expressions
Now, combine the simplified numerator and denominator:

\[
\frac{t\sqrt{1 + \frac{1}{t}}}{-t^2(1 - \frac{3}{t})} \approx \frac{t}{-t^2} = -\frac{1}{t}.
\]

### Step 4: Take the limit
As \(t \to \infty\), \(-\frac{1}{t} \to 0\).

Therefore, the limit is:

\[
\boxed{0}.
\]

Would you like a detailed explanation of any part of this process, or do you have any further questions?

Here are some related questions that might interest you:

1. How do we handle limits that involve infinity and indeterminate forms?
2. What are some common techniques for simplifying expressions under square roots?
3. How can we use factoring to simplify complex fractions?
4. How do we find limits involving rational functions at infinity?
5. What are other techniques for evaluating limits involving square roots?

**Tip:** Always look for the highest power of the variable in the numerator and denominator when simplifying limits that approach infinity.

Find the limit as t approaches infinity for (sqrt(t + t^2)) / (3t - t^2).

Explore how to find the limit of (sqrt(t + t^2)) / (3t - t^2) as t approaches infinity. Learn key concepts involving limits, rational functions, and simplifying expressions under square roots. This detailed solution is perfect for students at Grades 11-12 or introductory college-level calculus.

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