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.