Problem 56 - Thomas Calculus

Question

Suppose that \(\lim _{x \rightarrow-2} p(x)=4, \lim _{x \rightarrow-2} r(x)=0,\) and \(\lim _{x \rightarrow-2} s(x)=-3 .\) Find \begin{equation}\quad\ \text {a.} \lim _{x \rightarrow-2}(p(x)+r(x)+s(x)) \\\ \text {b.}\lim _{x \rightarrow-2} p(x) \cdot r(x) \cdot s(x) \\\ \quad\quad\quad\text {c.}\lim _{x \rightarrow-2}(-4 p(x)+5 r(x)) / s(x) \end{equation}

Step-by-Step Solution

Answer

a. 1 b. 0 c. \(\frac{16}{3}\)

Step 1: Identify known limits

Given limits are as follows; \(\lim_{x \to -2} p(x) = 4\), \(\lim_{x \to -2} r(x) = 0\), \(\lim_{x \to -2} s(x) = -3\). These will be used to simplify the expressions in parts (a), (b), and (c).

Step 2: Solution for part (a)

Apply the limit laws for sums: \(\lim_{x \to -2} (p(x) + r(x) + s(x)) = \lim_{x \to -2} p(x) + \lim_{x \to -2} r(x) + \lim_{x \to -2} s(x) = 4 + 0 - 3 = 1.\)

Step 3: Solution for part (b)

Apply the limit laws for products: \[\lim_{x \to -2} (p(x) \cdot r(x) \cdot s(x)) = \lim_{x \to -2} p(x) \cdot \lim_{x \to -2} r(x) \cdot \lim_{x \to -2} s(x) = 4 \cdot 0 \cdot (-3) = 0.\] The product is zero because \(\lim_{x \to -2} r(x) = 0\).

Step 4: Solution for part (c)

Use the limit law \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}\), provided \(\lim_{x \to a} g(x) eq 0\): \[\lim_{x \to -2} \frac{-4p(x) + 5r(x)}{s(x)} = \frac{-4 \cdot \lim_{x \to -2} p(x) + 5 \cdot \lim_{x \to -2} r(x)}{\lim_{x \to -2} s(x)} = \frac{-4 \cdot 4 + 5 \cdot 0}{-3} = \frac{-16}{-3} = \frac{16}{3}.\]

Key Concepts

Limit of a sum

When dealing with limits, an important concept is the **Limit of a Sum**, which tells us how to handle the limit of a combination of functions.
The Limit of a Sum rule states that if you have two functions, say \( u(x) \) and \( v(x) \), and you know their limits as \( x \) approaches a point, like

• \( \lim_{x \to a} u(x) = L_1 \)
• \( \lim_{x \to a} v(x) = L_2 \)

Then the limit of their sum is simply the sum of their limits:\[\lim_{x \to a} (u(x) + v(x)) = L_1 + L_2.\]This is exactly what was done in step 2 of the original exercise. By applying this rule, we plugged in the limits of the individual functions \( p(x), r(x), \) and \( s(x) \) to find that the limit of their sum was:\[\lim_{x \to -2} (p(x) + r(x) + s(x)) = 4 + 0 - 3 = 1.\]Using this rule can simplify solving limit problems by breaking down larger expressions into more manageable parts.

Limit of a product

The **Limit of a Product** is another important rule to grasp. It allows you to find the limit of a product of functions if you know the individual limits.
Suppose we have two functions \( f(x) \) and \( g(x) \) with known limits:

• \( \lim_{x \to a} f(x) = M \)
• \( \lim_{x \to a} g(x) = N \)

Then the product of these functions has a limit given by:\[\lim_{x \to a} (f(x) \cdot g(x)) = M \cdot N.\]In the exercise's step 3, the same rule was applied. Here, the expression \( p(x) \cdot r(x) \cdot s(x) \) was considered. By knowing their individual limits, the expression simplified to:\[\lim_{x \to -2} (p(x) \cdot r(x) \cdot s(x)) = 4 \cdot 0 \cdot (-3) = 0.\]Notice that if any limit is zero, such as \( r(x) \) in this case, the product will always be zero.

Limit of a quotient

Lastly, the **Limit of a Quotient** is an essential tool for solving limits involving division. Given two functions \( m(x) \) and \( n(x) \), the limit of their quotient involves

• \( \lim_{x \to a} m(x) = P \)
• \( \lim_{x \to a} n(x) = Q \)

The rule states that:\[\lim_{x \to a} \frac{m(x)}{n(x)} = \frac{P}{Q},\]provided \( Q eq 0 \). In the exercise, applying this rule to solve part (c) involves correctly identifying the limits of terms \(-4p(x) + 5r(x)\) and \(s(x)\).Important steps include:- Reviewing each limit: The numerator involves \(-4 \times \lim_{x \to -2} p(x) + 5 \times \lim_{x \to -2} r(x)\) which simplifies to \(-4 \times 4 + 5 \times 0 = -16\).- The denominator \( \lim_{x \to -2} s(x) = -3 \).The solution is finally:\[\lim_{x \to -2} \frac{-4p(x) + 5r(x)}{s(x)} = \frac{-16}{-3} = \frac{16}{3}.\]This demonstrates the application of the Limit of a Quotient rule, highlighting the necessity to ensure the denominator does not equal zero.