Math Problem Statement

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

Two graphs of piecewise functions. The upper is f(x), which has two linear segments. The first is a line with negative slope existing for x < -3. It goes toward the point (-3,0) at x= -3. The next has increasing slope and goes to the point (-3,-2) at x=-3. It exists for x > -3. Other key points are (0, 1), (-5,2), (1,2), (-7, 4), and (-9,6). The lower piecewise function has a linear segment and a curved segment. The linear segment exists for x < -3 and has decreasing slope. It goes to (-3,-2) at x=-3. The curved segment appears to be the right half of a downward opening parabola. It goes to the vertex point (-3,2) at x=-3. It crosses the y axis a little below y=-2. Other key points are (0, -7/3), (-5,0), (1,-5), (-7, 2), and (-9, 4). 118. limx→−3+(f(x)+g(x)) 119. limx→−3−(f(x)−3g(x)) 120. limx→0f(x)g(x)3 121. limx→−52+g(x)f(x) 122. limx→1(f(x))2 123. limx→1f(x)−g(x)−−−−−−−−−−√3 124. limx→−7(x⋅g(x)) 125. limx→−9[x⋅f(x)+2⋅g(x)]

Solution

To evaluate these limits, I'll analyze the given graphs and the limit laws. For clarity, I'll reference the values from the graphs at the specified xx-values.

118. lim⁡x→−3+(f(x)+g(x))\lim_{x \to -3^+} (f(x) + g(x))

From the right of x=−3x = -3:

  • f(x)→−2f(x) \to -2 (approaching from the right).
  • g(x)→2g(x) \to 2 (approaching from the right).

Thus: lim⁡x→−3+(f(x)+g(x))=−2+2=0.\lim_{x \to -3^+} (f(x) + g(x)) = -2 + 2 = 0.

119. lim⁡x→−3−(f(x)−3g(x))\lim_{x \to -3^-} (f(x) - 3g(x))

From the left of x=−3x = -3:

  • f(x)→0f(x) \to 0 (approaching from the left).
  • g(x)→−2g(x) \to -2 (approaching from the left).

Thus: lim⁡x→−3−(f(x)−3g(x))=0−3(−2)=0+6=6.\lim_{x \to -3^-} (f(x) - 3g(x)) = 0 - 3(-2) = 0 + 6 = 6.

120. lim⁡x→0f(x)g(x)3\lim_{x \to 0} \frac{f(x)g(x)}{3}

At x=0x = 0:

  • f(0)=1f(0) = 1 (value of f(x)f(x) at x=0x = 0).
  • g(0)=−73g(0) = -\frac{7}{3} (value of g(x)g(x) at x=0x = 0).

Thus: lim⁡x→0f(x)g(x)3=(1)(−73)3=−79.\lim_{x \to 0} \frac{f(x)g(x)}{3} = \frac{(1)(-\frac{7}{3})}{3} = \frac{-7}{9}.

121. lim⁡x→−52g(x)f(x)\lim_{x \to -5^2} \frac{g(x)}{f(x)}

At x=−5x = -5:

  • f(−5)=2f(-5) = 2 (value of f(x)f(x) at x=−5x = -5).
  • g(−5)=0g(-5) = 0 (value of g(x)g(x) at x=−5x = -5).

Thus: lim⁡x→−52g(x)f(x)=02=0.\lim_{x \to -5^2} \frac{g(x)}{f(x)} = \frac{0}{2} = 0.

122. lim⁡x→1(f(x))2\lim_{x \to 1} (f(x))^2

At x=1x = 1:

  • f(1)=2f(1) = 2 (value of f(x)f(x) at x=1x = 1).

Thus: lim⁡x→1(f(x))2=22=4.\lim_{x \to 1} (f(x))^2 = 2^2 = 4.

123. lim⁡x→1f(x)−g(x)3\lim_{x \to 1} \sqrt[3]{f(x) - g(x)}

At x=1x = 1:

  • f(1)=2f(1) = 2 (value of f(x)f(x) at x=1x = 1).
  • g(1)=−5g(1) = -5 (value of g(x)g(x) at x=1x = 1).

Thus: [ \lim_{x \to 1} \sqrt[3]{f(x) - g(x)} = \sqrt[3]{2 - (-5)} = \sqrt[3]{7}. ]

124. lim⁡x→−7(x⋅g(x))\lim_{x \to -7} (x \cdot g(x))

At x=−7x = -7:

  • g(−7)=2g(-7) = 2 (value of g(x)g(x) at x=−7x = -7).

Thus: lim⁡x→−7(x⋅g(x))=(−7)(2)=−14.\lim_{x \to -7} (x \cdot g(x)) = (-7)(2) = -14.

125. lim⁡x→−9[x⋅f(x)+2⋅g(x)]\lim_{x \to -9} [x \cdot f(x) + 2 \cdot g(x)]

At x=−9x = -9:

  • f(−9)=6f(-9) = 6 (value of f(x)f(x) at x=−9x = -9).
  • g(−9)=4g(-9) = 4 (value of g(x)g(x) at x=−9x = -9).

Thus: lim⁡x→−9[x⋅f(x)+2⋅g(x)]=(−9)(6)+2(4)=−54+8=−46.\lim_{x \to -9} [x \cdot f(x) + 2 \cdot g(x)] = (-9)(6) + 2(4) = -54 + 8 = -46.

Summary of Results

  1. lim⁡x→−3+(f(x)+g(x))=0\lim_{x \to -3^+} (f(x) + g(x)) = 0.
  2. lim⁡x→−3−(f(x)−3g(x))=6\lim_{x \to -3^-} (f(x) - 3g(x)) = 6.
  3. lim⁡x→0f(x)g(x)3=−79\lim_{x \to 0} \frac{f(x)g(x)}{3} = \frac{-7}{9}.
  4. lim⁡x→−52g(x)f(x)=0\lim_{x \to -5^2} \frac{g(x)}{f(x)} = 0.
  5. lim⁡x→1(f(x))2=4\lim_{x \to 1} (f(x))^2 = 4.
  6. lim⁡x→1f(x)−g(x)3=73\lim_{x \to 1} \sqrt[3]{f(x) - g(x)} = \sqrt[3]{7}.
  7. lim⁡x→−7(x⋅g(x))=−14\lim_{x \to -7} (x \cdot g(x)) = -14.
  8. lim⁡x→−9[x⋅f(x)+2⋅g(x)]=−46\lim_{x \to -9} [x \cdot f(x) + 2 \cdot g(x)] = -46.

Let me know if you'd like details or have any questions! 😊

Related Questions

  1. How do the limits change if f(x)f(x) or g(x)g(x) is redefined at discontinuity points?
  2. What are the conditions for applying limit laws in piecewise functions?
  3. How does the limit of a product behave if one of the factors approaches infinity?
  4. How does a cube root limit behave compared to a square root in terms of negative values?
  5. Can the difference of two limits ever fail to exist, even if the individual limits do?

Tip

When working with piecewise functions, always check the left- and right-hand limits separately at the points of discontinuity!

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Math Problem Analysis

Mathematical Concepts

Limit Laws
Piecewise Functions
Calculus
Continuity
Algebra

Formulas

lim(x→a) [f(x) + g(x)]
lim(x→a) [f(x) - 3g(x)]
lim(x→a) [f(x)g(x)]
lim(x→a) [x ⋅ f(x)]
lim(x→a) [x ⋅ f(x) + 2 ⋅ g(x)]
lim(x→a) [f(x)]^2
lim(x→a) [sqrt[3]{f(x) - g(x)}]

Theorems

Limit Laws
Continuity Theorem

Suitable Grade Level

Grades 11-12 (Advanced Calculus)

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