From the image you uploaded, there are five limits to evaluate. Here are the solutions to each:

1. $\lim_{x \to -1} e^x$

   • The exponential function is continuous, and therefore, the limit as $x$ approaches any real number is simply the function evaluated at that number. So: $\lim_{x \to -1} e^x = e^{-1} = \frac{1}{e}$

2. $\lim_{x \to 0} 2^{x-1}$

   • The function $2^{x-1}$ is a transformation of the function $2^x$, shifted down by 1. Since the exponential function $2^x$ is continuous, we directly substitute $x = 0$: $\lim_{x \to 0} 2^{x-1} = 2^{0-1} = 2^{-1} = \frac{1}{2}$

3. $\lim_{x \to 1} \ln(x^2 + x - 1)$

   • First, substitute $x = 1$ into the function: $x^2 + x - 1 = 1^2 + 1 - 1 = 1$
   • The natural log function is continuous where it is defined (i.e., $\ln(x)$ for $x > 0$): $\lim_{x \to 1} \ln(x^2 + x - 1) = \ln(1) = 0$

4. $\lim_{x \to \frac{\pi}{4}} \sin\left(\frac{\pi}{4} - x\right)$

   • Since $\sin$ is a continuous function: $\lim_{x \to \frac{\pi}{4}} \sin\left(\frac{\pi}{4} - x\right) = \sin\left(\frac{\pi}{4} - \frac{\pi}{4}\right) = \sin(0) = 0$

5. $\lim_{x \to -\frac{\pi}{2}} \cos\left(x + \frac{\pi}{2}\right)$

   • This can be simplified by recognizing a trigonometric identity: $\cos\left(x + \frac{\pi}{2}\right) = -\sin(x)$
   • Therefore: $\lim_{x \to -\frac{\pi}{2}} \cos\left(x + \frac{\pi}{2}\right) = -\sin\left(-\frac{\pi}{2}\right) = -(-1) = 1$

Would you like more details or have any further questions about these solutions? Here are some follow-up points to consider:

1. What would be the impact on these limits if the approaching values were changed?
2. How do these specific functions behave around the limit points?
3. Can these solutions be verified using graphical approaches?
4. What happens to these functions outside the limits provided?
5. How can the concept of limits be used in real-world applications?

Tip: When evaluating limits, especially involving trigonometric functions, remember to check for identities that can simplify calculations.