Let's evaluate both limits presented in the image.

(a) $\lim_{{x \to +\infty}} \frac{4}{e^x - 3}$

1. As $x \to +\infty$, the exponential term $e^x$ grows extremely large, far outpacing the constant $-3$. So we approximate:

   $e^x - 3 \approx e^x \quad \text{for very large } x.$

2. Therefore, the expression simplifies to:

   $\lim_{{x \to +\infty}} \frac{4}{e^x - 3} \approx \lim_{{x \to +\infty}} \frac{4}{e^x}.$

3. Since $e^x \to +\infty$ as $x \to +\infty$, the fraction approaches:

   $\frac{4}{e^x} \to 0.$

Thus, the limit for part (a) is:

$\lim_{{x \to +\infty}} \frac{4}{e^x - 3} = 0.$

(b) $\lim_{{x \to -\infty}} \frac{4}{e^x - 3}$

1. As $x \to -\infty$, the exponential $e^x \to 0$, because $e^x$ decays to 0 for negative values of $x$. So we have:

   $e^x - 3 \approx -3 \quad \text{for very negative } x.$

2. Therefore, the limit becomes:

   $\lim_{{x \to -\infty}} \frac{4}{e^x - 3} \approx \lim_{{x \to -\infty}} \frac{4}{-3}.$

3. Simplifying:

   $\frac{4}{-3} = -\frac{4}{3}.$

Thus, the limit for part (b) is:

$\lim_{{x \to -\infty}} \frac{4}{e^x - 3} = -\frac{4}{3}.$

If you have any questions or would like further explanation, feel free to ask!

Here are some related questions to expand your understanding:

1. What is the general behavior of exponential functions as $x \to +\infty$ and $x \to -\infty$?
2. How does adding or subtracting constants affect the behavior of limits involving exponentials?
3. Can you apply L'Hôpital's Rule to evaluate limits involving exponentials? Why or why not in this case?
4. What other common functions behave similarly to exponentials at infinity?
5. How would the limit change if the exponent were negative, i.e., $e^{-x}$?

Tip: Exponential functions grow faster than polynomials and constant terms as $x \to +\infty$, which is helpful for evaluating limits.