https://file.aiquickdraw.com/mathbot/file/1725957780274v2uglshv.jpg

The expression in the image is:

\[
\lim_{{n \to \infty}} \left( \frac{n - e^{2n}}{n} \right)
\]

Let's break this down and solve the limit:

### Expression:
\[
\lim_{{n \to \infty}} \left( \frac{n - e^{2n}}{n} \right)
\]

We can rewrite this as:
\[
\lim_{{n \to \infty}} \left( 1 - \frac{e^{2n}}{n} \right)
\]

### Analyzing the limit:

- As \( n \to \infty \), \( e^{2n} \) grows exponentially, much faster than \( n \), meaning \( \frac{e^{2n}}{n} \to \infty \).
- Thus, the whole expression \( 1 - \frac{e^{2n}}{n} \to -\infty \).

### Conclusion:

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

Let me know if you need further clarification on this limit.

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Would you like to dive into further details or have additional questions? Here are five related questions to explore:

1. How does the behavior of exponential functions like \( e^{2n} \) compare to polynomial growth as \( n \to \infty \)?
2. Can we use L'Hopital's Rule to solve a similar limit with a different form?
3. How does the growth rate of \( e^n \) compare to other common functions like \( \log(n) \) or \( n^2 \)?
4. What happens if we modify the power in the exponential function, for example \( e^n \) instead of \( e^{2n} \)?
5. How would the result change if the denominator were \( n^2 \) instead of \( n \)?

**Tip:** When analyzing limits, comparing the growth rates of the terms involved (polynomial, logarithmic, exponential) can provide insight into the behavior of the overall function.

What is the limit of (n - e^(2n)) / n as n approaches infinity?

Learn how to evaluate the limit lim_(n → ∞) ((n - e^(2n)) / n), where exponential growth dominates over polynomial terms. Understand the behavior of the function as n approaches infinity and why the limit results in negative infinity.

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