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We are tasked with determining the limit:

\[
\lim_{x \to -\infty} \left( 6 \exp(-x + 8) + 2 \right)
\]

### Solution:

First, we simplify the exponential term:

\[
\exp(-x + 8) = \exp(-x) \cdot \exp(8)
\]

Since \(\exp(8)\) is just a constant, we now focus on the behavior of \(\exp(-x)\) as \(x \to -\infty\).

- When \(x \to -\infty\), \(\exp(-x)\) behaves as follows:
  \[
  \exp(-x) = e^{-x} \to e^{+\infty} = \infty
  \]

Thus, the entire expression for the exponential grows without bound as \(x \to -\infty\). Therefore:

\[
6 \exp(-x + 8) \to 6 \cdot \infty = \infty
\]

Finally, since the added constant \(2\) does not affect the infinity, the entire limit becomes:

\[
\lim_{x \to -\infty} \left( 6 \exp(-x + 8) + 2 \right) = \infty
\]

Thus, the limit is \(\infty\).

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How does the behavior of \(\exp(x)\) change as \(x \to \infty\) compared to \(x \to -\infty\)?
2. What is the limit of \(6 \exp(-x + 8)\) as \(x \to \infty\)?
3. How does adding or subtracting constants inside the exponential affect the function's growth?
4. How do limits involving exponential functions differ from polynomial limits as \(x \to \infty\)?
5. Can you think of other functions where limits as \(x \to -\infty\) behave similarly?

**Tip**: When evaluating limits at infinity, always check the dominant term. In this case, the exponential grows much faster than any constant term.

Determine the limit: lim (x → -∞) [6 exp(-x + 8) + 2]

Learn how to determine the limit of the function 6 exp(-x + 8) + 2 as x approaches negative infinity. This problem involves the properties of exponential functions and their behavior at infinity. Suitable for advanced high school or pre-university students.

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