Evaluate the given limits using the graph of the function f(x)=−1(x−4)(x−9)2 shown above. 
Enter Infinity for ∞, -Infinity for −∞, or DNE if the limit does not exist, but is neither ∞ nor −∞.
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a) limx→4−f(x) =  
b) limx→4+f(x) =  
c) limx→4f(x) =  
d) limx→9−f(x) =  
e) limx→9+f(x) =  
f) limx→9f(x) = 


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.

Evaluate the given limits using the graph of the function f(x)=−1(x−4)(x−9)² shown above. Enter Infinity for ∞, -Infinity for −∞, or DNE if the limit does not exist, but is neither ∞ nor −∞. a) limx→4−f(x) = b) limx→4+f(x) = c) limx→4f(x) = d) limx→9−f(x) = e) limx→9+f(x) = f) limx→9f(x) = 

Learn how to evaluate limits for the function f(x) = -1 / ((x - 4) * (x - 9)²) near points x = 4 and x = 9. This solution involves determining one-sided and two-sided limits, as well as analyzing asymptotic behavior at these points. Suitable for high school or college calculus students.

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