The image provided contains two limit problems to be evaluated. Let's analyze and solve each one:

Problem (a)

$\lim_{x \to \infty} \frac{2}{e^x - 7}$ As $x \to \infty$, the exponential function $e^x$ grows very rapidly. Therefore, the term $e^x - 7$ will also grow very large because subtracting 7 from a very large number will not change its magnitude significantly.

Thus, we can approximate: $e^x - 7 \approx e^x \quad \text{as} \quad x \to \infty$ So, the limit becomes: $\lim_{x \to \infty} \frac{2}{e^x} = 0$ Therefore, the solution to (a) is: $\boxed{0}$

Problem (b)

$\lim_{x \to -\infty} \frac{2}{e^x - 7}$ As $x \to -\infty$, the exponential function $e^x$ approaches 0, because $e^x$ decays rapidly as $x$ becomes more negative. Therefore, $e^x - 7$ approaches: $e^x - 7 \approx -7 \quad \text{as} \quad x \to -\infty$ So the limit becomes: $\lim_{x \to -\infty} \frac{2}{-7} = -\frac{2}{7}$ Thus, the solution to (b) is: $\boxed{-\frac{2}{7}}$

Summary of Results:

• (a) $\lim_{x \to \infty} \frac{2}{e^x - 7} = 0$
• (b) $\lim_{x \to -\infty} \frac{2}{e^x - 7} = -\frac{2}{7}$

Would you like a detailed explanation on any of the steps?

Here are 5 related questions you might find helpful:

1. What happens if the function inside the limit is polynomial instead of exponential?
2. How does the behavior of exponential functions compare to logarithmic functions as $x \to \infty$?
3. What are the properties of the exponential function $e^x$ for large negative $x$?
4. How can L'Hopital's Rule be used for evaluating limits involving indeterminate forms?
5. How do horizontal asymptotes relate to limits as $x$ approaches infinity?

Tip: When solving limits involving exponentials, always remember the rapid growth rate of $e^x$ as $x \to \infty$. It helps simplify complex expressions.