Let's evaluate the limits step by step:

1. $\lim_{x \to 3^+} \frac{2}{x - 3}$

This is a one-sided limit as $x$ approaches $3$ from the right ($x \to 3^+$).

• As $x \to 3^+$, $x - 3$ approaches $0$ from the positive side, meaning $x - 3 > 0$.
• Therefore, the fraction $\frac{2}{x - 3}$ grows without bound, and the limit is positive infinity.

$\lim_{x \to 3^+} \frac{2}{x - 3} = \infty$

2. $\lim_{x \to 5} \frac{2}{(x - 5)^3}$

Here, $x$ is approaching 5.

• As $x \to 5$, $x - 5$ approaches 0. The cube, $(x - 5)^3$, will be a very small number but retains the sign of $x - 5$.

• Since the cube of a small negative number is still negative and of a small positive number is positive, the limit will depend on whether we approach 5 from the right or left.

  • If $x \to 5^+$ (from the right), $(x - 5)^3 > 0$ and hence $\frac{2}{(x - 5)^3} \to \infty$.
  • If $x \to 5^-$ (from the left), $(x - 5)^3 < 0$ and hence $\frac{2}{(x - 5)^3} \to -\infty$.

So the overall limit does not exist because the left and right limits do not agree. Thus, the limit is:

$\lim_{x \to 5} \frac{2}{(x - 5)^3} = \text{DNE (Does Not Exist)}$

3. $\lim_{x \to -7} \frac{1}{x^2(x + 7)}$

• As $x \to -7$, notice that $x + 7 \to 0$.

• The term $x^2$ does not approach 0 since $x^2$ is always positive for any real $x$, and at $x = -7$, $x^2 = 49$.

• So, the function behaves like $\frac{1}{49 \cdot 0}$, which suggests that the limit is infinite.

• More precisely, as $x \to -7$, $x + 7$ approaches 0 from either side, but since $x^2$ is positive, the sign of the result is determined by $x + 7$:

  • From the left ($x \to -7^-$), $x + 7$ is negative, so the limit approaches $-\infty$.
  • From the right ($x \to -7^+$), $x + 7$ is positive, so the limit approaches $+\infty$.

Thus, the limit does not exist because of the differing signs.

$\lim_{x \to -7} \frac{1}{x^2(x + 7)} = \text{DNE (Does Not Exist)}$

4. $\lim_{x \to 5} \frac{2}{(x - 5)^6}$

• As $x \to 5$, $(x - 5)^6$ approaches 0.
• Since $(x - 5)^6$ is always positive (even for values close to 5), the fraction $\frac{2}{(x - 5)^6}$ tends to $\infty$.

$\lim_{x \to 5} \frac{2}{(x - 5)^6} = \infty$

Summary of Results:

1. $\infty$
2. $\text{DNE}$
3. $\text{DNE}$
4. $\infty$

Would you like further clarification or a more detailed explanation on any of these? Here are five questions to expand on this topic:

1. How do one-sided limits differ from two-sided limits?
2. What are common techniques to evaluate limits involving infinity?
3. How does the behavior of even and odd powers of a variable near 0 affect limits?
4. Can limits that result in infinity be classified as existing or not existing?
5. How would adding or multiplying terms change the evaluation of these limits?

Tip: When dealing with limits approaching a value where the denominator tends to zero, always consider the sign of the terms near the limit to determine the behavior (positive or negative infinity).