Let's solve each of these limit problems individually, as well as identify the location of the vertical asymptote.

1. First limit: $\lim_{{x \to 2^-}} \frac{(x+3)^2}{(x-2)^3}$

   As $x$ approaches 2 from the left, the expression $(x-2)$ approaches 0 from the negative side, so $(x-2)^3$ approaches 0 from the negative side as well. Since the numerator $(x+3)^2$ remains positive (it approaches $5^2 = 25$), the entire fraction will approach $-\infty$ as $x \to 2^-$.

   Therefore: $\lim_{{x \to 2^-}} \frac{(x+3)^2}{(x-2)^3} = -\infty$

2. Second limit: $\lim_{{x \to 2^+}} \frac{(x+3)^2}{(x-2)^3}$

   Here, as $x$ approaches 2 from the right, $(x-2)$ approaches 0 from the positive side, making $(x-2)^3$ approach 0 from the positive side as well. Since $(x+3)^2$ remains positive (again approaching 25), the entire fraction will approach $+\infty$ as $x \to 2^+$.

   Thus: $\lim_{{x \to 2^+}} \frac{(x+3)^2}{(x-2)^3} = +\infty$

3. Vertical Asymptote: The expression $\frac{(x+3)^2}{(x-2)^3}$ has a vertical asymptote at $x = 2$, where the denominator $(x-2)^3 = 0$.

   Therefore, the answers are:

   • $\lim_{{x \to 2^-}} \frac{(x+3)^2}{(x-2)^3} = -\infty$
   • $\lim_{{x \to 2^+}} \frac{(x+3)^2}{(x-2)^3} = +\infty$
   • Vertical asymptote at $x = 2$.

Would you like further details on any of these steps or have any questions?

Here are some follow-up questions to deepen understanding:

1. What are left-hand and right-hand limits, and why do they matter in this problem?
2. How would the limits change if the exponent of $(x-2)$ in the denominator were even?
3. What is the difference between a vertical asymptote and a hole in the graph of a function?
4. Can limits approaching $+\infty$ or $-\infty$ ever indicate a removable discontinuity?
5. How do the signs of $(x-2)$ influence the behavior of the fraction as $x \to 2$?

Tip: When evaluating limits involving powers of terms approaching zero, always consider the direction of approach to determine the sign of the resulting expression.