To solve these limit problems for the function $f(x) = -\frac{1}{(x-4)(x-9)^2}$, we will analyze the behavior of the function near the points $x = 4$ and $x = 9$.

Given function:

$f(x) = -\frac{1}{(x-4)(x-9)^2}$

Let's break down each limit:

Part (a) $\lim_{x \to 4^-} f(x)$

• As $x \to 4^-$, the factor $(x - 4)$ approaches 0 from the left, making it negative.
• The term $(x - 9)^2$ remains positive, as squaring any real number results in a positive value.
• Thus, near $x = 4^-$, the function becomes very negative because of the negative sign in front of the fraction and the negative $(x-4)$ term.

So, $\lim_{x \to 4^-} f(x) = -\infty$

Part (b) $\lim_{x \to 4^+} f(x)$

• As $x \to 4^+$, the factor $(x - 4)$ approaches 0 from the right, making it positive.
• The term $(x - 9)^2$ is still positive.
• Therefore, the overall expression becomes positive since both $(x-4)$ and $(x-9)^2$ are positive, but with a negative sign in front of the fraction.

So, $\lim_{x \to 4^+} f(x) = \infty$

Part (c) $\lim_{x \to 4} f(x)$

• Since the left-hand and right-hand limits as $x \to 4$ approach $-\infty$ and $\infty$, respectively, the overall limit does not exist.

So, $\lim_{x \to 4} f(x) = \text{DNE}$

Part (d) $\lim_{x \to 9^-} f(x)$

• As $x \to 9^-$, the factor $(x - 9)^2$ becomes very small but positive (since it's squared).
• The factor $(x - 4)$ remains positive.
• Therefore, the function approaches $-\infty$ due to the negative sign in front of the fraction and the small value of $(x - 9)^2$.

So, $\lim_{x \to 9^-} f(x) = -\infty$

Part (e) $\lim_{x \to 9^+} f(x)$

• As $x \to 9^+$, the same analysis applies. The factor $(x - 9)^2$ is still positive and small, and $(x - 4)$ remains positive.
• So the behavior is the same as for $x \to 9^-$, and the function again approaches $-\infty$.

So, $\lim_{x \to 9^+} f(x) = -\infty$

Part (f) $\lim_{x \to 9} f(x)$

• Since both the left-hand and right-hand limits as $x \to 9$ approach $-\infty$, the overall limit is also $-\infty$.

So, $\lim_{x \to 9} f(x) = -\infty$

Final Answers:

a) $\lim_{x \to 4^-} f(x) = -\infty$

b) $\lim_{x \to 4^+} f(x) = \infty$

c) $\lim_{x \to 4} f(x) = \text{DNE}$

d) $\lim_{x \to 9^-} f(x) = -\infty$

e) $\lim_{x \to 9^+} f(x) = -\infty$

f) $\lim_{x \to 9} f(x) = -\infty$

Would you like any further clarification or details on these calculations?

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Related Questions:

1. What happens to the function as $x \to -\infty$?
2. How would the behavior change if $(x - 9)^2$ was not squared?
3. Can you explain why the limit does not exist at $x = 4$?
4. How would the function behave if the negative sign in the numerator were removed?
5. What are the critical points and asymptotes of this function?

Tip:

When evaluating limits near discontinuities or singularities, consider analyzing one-sided limits to determine overall behavior.