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Calculus IChapter 2: LimitsSection 2-4

Limits Properties — Question 6

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Question 6

Let f(x)f(x) and g(x)g(x) be functions such that: limx→2f(x)=5,limx→2g(x)=−3\lim_{x \to 2} f(x) = 5, \quad \lim_{x \to 2} g(x) = -3

Use limit laws to compute the following:

(a) lim⁡x→2[3f(x)−2g(x)]\displaystyle \lim_{x \to 2} [3f(x) - 2g(x)]

(b) lim⁡x→2[f(x)g(x)]\displaystyle \lim_{x \to 2} \left[ \frac{f(x)}{g(x)} \right]

Original worksheet page 1: question and worked solution for 2-4-006
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Question 6 - Solution

We are given:

limx→2f(x)=5,limx→2g(x)=−3\lim_{x \to 2} f(x) = 5, \quad \lim_{x \to 2} g(x) = -3

(a) Use limit laws for scalar multiplication and addition/subtraction:

limx→2[3f(x)−2g(x)]=3⋅limx→2f(x)−2⋅limx→2g(x)=3(5)−2(−3)=15+6=21\begin{align*} \lim_{x \to 2} [3f(x) - 2g(x)] &= 3 \cdot \lim_{x \to 2} f(x) - 2 \cdot \lim_{x \to 2} g(x) \\ &= 3(5) - 2(-3) = 15 + 6 = \boxed{21} \end{align*}

(b) Use the quotient limit law:

limx→2[f(x)g(x)]=limx→2f(x)limx→2g(x)=5−3=−53\lim_{x \to 2} \left[ \frac{f(x)}{g(x)} \right] = \frac{\lim_{x \to 2} f(x)}{\lim_{x \to 2} g(x)} = \frac{5}{-3} = \boxed{-\frac{5}{3}}

Original worksheet page 2: question and worked solution for 2-4-006

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