Determine lim(square root (9n^2+8n+12)—3n+18

We need to find:

\[
\lim_{n \to \infty} \left(\sqrt{9n^2 + 8n + 12} - 3n + 18\right)
\]

### Step 1: Simplify the expression
First, focus on simplifying \(\sqrt{9n^2 + 8n + 12} - 3n\) because it's the part that depends on \(n\). 

Factor out \(n^2\) from inside the square root:

\[
\sqrt{9n^2 + 8n + 12} = \sqrt{n^2(9 + \frac{8}{n} + \frac{12}{n^2})} = n\sqrt{9 + \frac{8}{n} + \frac{12}{n^2}}
\]

Thus, the expression becomes:

\[
\sqrt{9n^2 + 8n + 12} - 3n = n\sqrt{9 + \frac{8}{n} + \frac{12}{n^2}} - 3n
\]

### Step 2: Use the approximation for large \(n\)
As \(n \to \infty\), the terms \(\frac{8}{n}\) and \(\frac{12}{n^2}\) approach 0. Therefore, for large \(n\), we can approximate:

\[
\sqrt{9 + \frac{8}{n} + \frac{12}{n^2}} \approx \sqrt{9} = 3
\]

This leads to:

\[
n\sqrt{9 + \frac{8}{n} + \frac{12}{n^2}} - 3n \approx n(3) - 3n = 0
\]

### Step 3: Refine the approximation
For a more accurate result, expand \(\sqrt{9 + \frac{8}{n} + \frac{12}{n^2}}\) using a binomial expansion around infinity. The binomial expansion of \(\sqrt{1+x}\) for small \(x\) is:

\[
\sqrt{1 + x} \approx 1 + \frac{x}{2} \quad \text{for small } x
\]

In our case, we have:

\[
\sqrt{9 + \frac{8}{n} + \frac{12}{n^2}} = 3\sqrt{1 + \frac{8}{9n} + \frac{12}{9n^2}}
\]

For large \(n\), we expand the square root:

\[
\approx 3\left(1 + \frac{4}{9n}\right)
\]

Thus:

\[
n\sqrt{9 + \frac{8}{n} + \frac{12}{n^2}} - 3n \approx n\left(3 + \frac{4}{n}\right) - 3n = n(3) + 4 - 3n = 4
\]

### Step 4: Final expression
Now, adding back the constant 18:

\[
\lim_{n \to \infty} \left(\sqrt{9n^2 + 8n + 12} - 3n + 18\right) = 4 + 18 = 22
\]

### Conclusion:
\[
\lim_{n \to \infty} \left(\sqrt{9n^2 + 8n + 12} - 3n + 18\right) = 22
\]

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Would you like more details or have any questions?

Here are 5 related questions to extend your understanding:
1. How does binomial expansion help simplify square roots for large values of \(n\)?
2. What is the behavior of polynomials of different degrees as \(n \to \infty\)?
3. How can you apply limits to other functions that involve square roots?
4. What happens when higher-order terms are included in approximations for limits?
5. How would the limit change if the coefficients in the original expression were different?

**Tip:** When dealing with large values in limits, approximating small terms or using expansions can significantly simplify complex expressions.

Determine lim(square root(9n^2+8n+12) - 3n + 18)

Explore the limit of sqrt(9n^2 + 8n + 12) - 3n + 18 as n approaches infinity. This problem involves applying limits, binomial expansion, and asymptotic approximations to simplify the expression. Suitable for advanced high school or early college students.

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