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what is the limit of the attached equation

The given expression in the image is:

\[
\left( \frac{n^2 + n + 1}{n^2 + n + 2} \right)^{2n^2 + n + 1}
\]

To find the limit as \( n \to \infty \), we can analyze the behavior of both the base and the exponent separately:

### Step 1: Behavior of the base
The base is:

\[
\frac{n^2 + n + 1}{n^2 + n + 2}
\]

For large \( n \), the dominant terms in both the numerator and the denominator are \( n^2 \). Therefore, as \( n \to \infty \), we can approximate the base as:

\[
\frac{n^2 + n + 1}{n^2 + n + 2} \approx \frac{n^2}{n^2} = 1
\]

Thus, the base tends to 1 as \( n \) becomes large.

### Step 2: Exponent behavior
The exponent is \( 2n^2 + n + 1 \). As \( n \to \infty \), this grows very quickly, much faster than the base approaches 1.

### Step 3: Applying the limit
The expression is of the form \( (1 + \text{small term})^{\text{large term}} \). To evaluate such limits, we can use the fact that:

\[
\lim_{x \to 0} \left( 1 + x \right)^{1/x} = e
\]

We rewrite the base as:

\[
\frac{n^2 + n + 1}{n^2 + n + 2} = 1 - \frac{1}{n^2 + n + 2}
\]

Thus, for large \( n \), the base becomes approximately \( 1 - \frac{1}{n^2} \), and the whole expression can be written as:

\[
\left( 1 - \frac{1}{n^2} \right)^{2n^2 + n + 1}
\]

Now, approximate this using logarithms:

\[
\lim_{n \to \infty} \left( 1 - \frac{1}{n^2} \right)^{2n^2} \approx \exp\left( -\frac{2n^2}{n^2} \right) = \exp(-2)
\]

### Final Answer:
The limit of the expression as \( n \to \infty \) is:

\[
e^{-2}
\]

Let me know if you would like a more detailed explanation of any step!

### Related questions:
1. What happens if the exponent grows even faster?
2. How would the result change if the base had a different leading term in the numerator?
3. Can this problem be solved using L'Hopital's Rule?
4. How do small changes in the base affect the overall limit?
5. How would the limit behave for different forms of polynomial growth in the exponent?

### Tip:
When analyzing limits involving exponents and bases that tend to 1, consider using logarithmic approximations to simplify the process.

What is the limit of the attached equation?

Explore the limit of an algebraic fraction raised to a rapidly growing exponent. This problem uses limits and logarithmic approximations to find that the final limit is e^-2 as n tends to infinity. Suitable for high school students in Grades 11-12 studying advanced calculus.

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