Name Math 109 Quiz 6 Section1 Determine The Following Limitsa Li ✓ Solved

Name Math 109 - Quiz 6 Section 1. Determine the following limits: (a) lim x→3 2x2−x+7 2x−3 (b) limx→π cos x 2−sin x (c) limx→π2 x cos2 x 1−sin x 2. A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides. Given 131 ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?

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Evaluate the following limits. (a) lim x→∞ 3x2−6x5 2x5+3x2−1 (b) lim x→∞ 3x √ x2+1 (c) lim x→−∞ 3x √ x2+1 For use with OpenStax Calculus, free at 54 Name: Limits and Asymptotes Section: 3. Evaluate the following limits. (a) lim x→∞ sin x+cos x sin x−cos x (b) limx→∞ 10 cos x x (c) limx→∞ 1−5ex ex 4. Evaluate the following limits. (a) lim x→−∞ 3x5+x4 100x4−1 (b) lim x→−∞ x2 √ 1−x (c) lim x→−∞ 101x ln(−x)+20x2 4x2+1 5. If f ′(x) has asymptotes at y = 3 and x = 1, then f (x) has what asymptotes? Sketch a possible graph for f (x) and f ′(x) on the same set of axes.

6. Determine the oblique asymptotes of f (x) = √ x2 − 5x + 4 For use with OpenStax Calculus, free at 55 Name: L’Hopital’s Rule Section: 4.8 L’Hopital’s Rule Vocabulary Examples L’Hopital’s Rule f and g are differentiable functions over an open interval containing a. If lim x→a f (x) = and lim x→a g(x) = , then lim x→a f (x) g(x) = In addition, if lim x→a f (x) = and lim x→a g(x) = , then lim x→a f (x) g(x) = 1. Determine whether L’Hopital’s Rule can be applied to each of the following limits. (a) lim x→∞ ex x (b) lim x→∞ x ln x (c) lim x→1 x2 x−2 (d) lim x→0 1−cos x x Determine the following limits. 2. lim x→0 sin x x 3. limx→1 ln x x−1 4. limx→∞ ex x2 5. lim x→π sin x x 6. limx→0 1−cos x x 7. limx→−∞ cos(ex) ex For use with OpenStax Calculus, free at 60 Name: L’Hopital’s Rule Section: Determine the following limits.

8. lim x→π π−x sin x 9. limx→π2 sec x 1+tan x 10. lim x→3 xex x−. lim x→+x)n−1 x 12. lim x→0 sin x−tan x x. lim x→∞ ln x x. lim x→0 √ 1+x− √ 1−x x 15. lim x→0 ex−x−1 x2 16. lim x→0 tan x √ x 17. Evaluate the following limits for a , 0 (a) lim x→a x−a x2−a2 (b) lim x→a x−a x3−a3 (c) lim x→a x−a xn−an For use with OpenStax Calculus, free at 61

Paper for above instructions

Assignment Solution: Limits and Maximum Area

Part 1: Calculate the Limits

In this section, we evaluate the limits provided in the quiz:
(a) \( \lim_{x \to 3} \frac{2x^2 - x + 7}{2x - 3} \)
To find this limit, we substitute \( x = 3 \):
\[
\frac{2(3)^2 - 3 + 7}{2(3) - 3} = \frac{2(9) - 3 + 7}{6 - 3} = \frac{18 - 3 + 7}{3} = \frac{22}{3}
\]
Therefore,
\[
\lim_{x \to 3} \frac{2x^2 - x + 7}{2x - 3} = \frac{22}{3}
\]
(b) \( \lim_{x \to \pi} \frac{\cos x}{2 - \sin x} \)
Substituting \( x = \pi \):
\[
\frac{\cos(\pi)}{2 - \sin(\pi)} = \frac{-1}{2 - 0} = \frac{-1}{2}
\]
Thus,
\[
\lim_{x \to \pi} \frac{\cos x}{2 - \sin x} = -\frac{1}{2}
\]
(c) \( \lim_{x \to \pi^2} \frac{x \cos^2 x}{1 - \sin x} \)
Plugging in \( x = \pi^2 \):
\[
\frac{\pi^2 \cos^2(\pi^2)}{1 - \sin(\pi^2)}
\]
Here, \( \cos^2(\pi^2) \) and \( \sin(\pi^2) \) need values:
- \( \cos(\pi^2) \) and \( \sin(\pi^2) \) approximate to \(-0.416\) and \(0.412\) respectively.
Calculating gives us:
\[
\frac{\pi^2 (-0.416)^2}{1 - 0.412} = \frac{\pi^2 (0.173056)}{0.588} \approx \frac{0.544}{0.588} \approx 0.925
\]
Therefore,
\[
\lim_{x \to \pi^2} \frac{x \cos^2 x}{1 - \sin x} \approx 0.925
\]

Part 2: Maximum Area of Rectangular Garden

To find the dimensions that create a garden of maximum area with three sides made of wire fencing and one side a rock wall, we can denote:
- Let \( x \) be the width parallel to the rock wall.
- Let \( y \) be the length perpendicular to the rock wall.
The perimeter constraint can be expressed as:
\[
x + 2y = 131
\]
From this equation, we can express \( y \) in terms of \( x \):
\[
2y = 131 - x \quad \Rightarrow \quad y = \frac{131 - x}{2}
\]
The area \( A \) of the garden can be represented as:
\[
A = x \cdot y = x \cdot \frac{131 - x}{2} = \frac{131x - x^2}{2}
\]
To maximize the area, we take the derivative of \( A \) with respect to \( x \) and set it to zero:
\[
A' = \frac{1}{2} (131 - 2x) = 0
\]
Solving for \( x \):
\[
131 - 2x = 0 \quad \Rightarrow \quad 2x = 131 \quad \Rightarrow \quad x = \frac{131}{2} = 65.5 \text{ ft}
\]
Substituting back to find \( y \):
\[
y = \frac{131 - 65.5}{2} = \frac{65.5}{2} = 32.75 \text{ ft}
\]
The maximum area \( A \):
\[
A = 65.5 \cdot 32.75 \approx 2141.125 \text{ square feet}
\]

Conclusion

The maximum area of the rectangular garden is approximately \( 2141.125 \) square feet, with dimensions:
- Width (along the rock wall): \( 65.5 \) ft
- Length: \( 32.75 \) ft

References

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