Let’s evaluate each limit step by step.

1. $\lim_{x \to 5} \frac{2}{(x - 5)^6}$
At $x = 5$, the denominator becomes $(5 - 5)^6 = 0^6 = 0$. This makes the fraction undefined.
To determine the behavior near $x = 5$:

• When $x$ approaches 5 from the right ($x \to 5^+$), $(x - 5)^6 > 0$. Since $2/(x - 5)^6 > 0$, the limit approaches $+\infty$.
• When $x$ approaches 5 from the left ($x \to 5^-$), $(x - 5)^6 > 0$ (because $n = 6$ is even). The value still approaches $+\infty$.

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

2. $\lim_{x \to 3} \frac{2}{x - 3}$
Here, the denominator becomes $x - 3 = 0$ at $x = 3$. Again, we analyze the behavior near $x = 3$:

• When $x \to 3^+$, $x - 3 > 0$. Hence $\frac{2}{x - 3} \to +\infty$.
• When $x \to 3^-$, $x - 3 < 0$. Hence $\frac{2}{x - 3} \to -\infty$.

Since the left- and right-hand limits are not the same, the limit does not exist.

3. $\lim_{x \to 0} \frac{1}{x^2(x + 7)}$
At $x = 0$, the denominator becomes $x^2(x + 7) = 0$. Analyzing the behavior:

• $x^2 > 0$ for all $x \neq 0$, and $x + 7 > 0$ for all $x$ near 0.
• Thus, the denominator approaches $0^+$, making $\frac{1}{x^2(x + 7)} \to +\infty$.

Therefore:
$\lim_{x \to 0} \frac{1}{x^2(x + 7)} = \infty$

4. $\lim_{x \to -7} \frac{1}{x^2(x + 7)}$
At $x = -7$, the denominator becomes $x^2(x + 7) = (-7)^2(-7 + 7) = 49(0) = 0$.
Near $x = -7$, consider the sign of $x + 7$:

• When $x \to -7^+$, $x + 7 > 0$, so the denominator is $x^2(x + 7) > 0$. Thus, $\frac{1}{x^2(x + 7)} \to +\infty$.
• When $x \to -7^-$, $x + 7 < 0$, so the denominator is $x^2(x + 7) < 0$. Thus, $\frac{1}{x^2(x + 7)} \to -\infty$.

Since the left- and right-hand limits are not the same, the limit does not exist.

Final Answers:

1. $\infty$
2. Does not exist
3. $\infty$
4. Does not exist

Would you like further clarifications or detailed steps for any part of the solution?

Related Questions:

1. What does it mean for a limit to approach infinity, and how is it interpreted in graphs?
2. How do we determine one-sided limits for rational functions with vertical asymptotes?
3. Why does an even exponent in the denominator (e.g., $(x - 5)^6$) affect the sign of the limit?
4. What happens to limits if the numerator and denominator both approach zero (indeterminate forms)?
5. How can limits involving infinity help us describe the behavior of functions at their vertical asymptotes?

Tip:

When analyzing limits with denominators approaching zero, always check the sign of the denominator near the critical point to determine the direction of the behavior (positive or negative infinity).