Problem 50 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises 49 and 50 , determine the limit. Assume that $$\lim _ { x \rightarrow b } f ( x ) = 7$$ and $$\lim _ { x \rightarrow b } g ( x ) = - 3$$ (a) $$\lim _ { x \rightarrow b } ( f ( x ) + g ( x ) ) \quad \quad$$ (b) $$\lim _ { x \rightarrow b } ( f ( x ) \cdot g ( x ) )$$ (c) $$\lim _ { x \rightarrow b } 4 g ( x ) - \quad$$ (d) $$\lim _ { x \rightarrow b } \frac { f ( x ) } { g ( x ) }$$

Step-by-Step Solution

Answer

(a) The limit is 4. (b) The limit is -21. (c) The limit is -12. (d) The limit is \(-\frac{7}{3}\).

Step 1: Find the limit of (f(x) + g(x))

The sum of the limits of two functions is equal to the limit of their sum. Therefore, we substitute the given limits into the sum, like so: \(\lim _ { x \rightarrow b } ( f ( x ) + g ( x ) ) = \lim _ { x \rightarrow b } f ( x ) + \lim _ { x \rightarrow b } g ( x ) = 7 - 3 = 4\)

Step 2: Find the limit of (f(x) * g(x))

The limit of the product of two functions is equal to the product of their limits. We substitute the given limits into the sum like so: \(\lim _ { x \rightarrow b } ( f ( x ) \cdot g ( x ) ) = \lim _ { x \rightarrow b } f ( x ) \cdot \lim _ { x \rightarrow b } g ( x ) = 7 \cdot -3 = -21\)

Step 3: Find the limit of 4g(x)

The limit of a constant times a function is equal to the constant times the limit of the function. We substitute the given limit of g(x) into the equation like: \(\lim _ { x \rightarrow b } 4 g ( x ) = 4 \cdot \lim _ { x \rightarrow b } g ( x ) = 4 \cdot -3 = -12\)

Step 4: Find the limit of (f(x) / g(x))

The limit of the division of two functions is equal to the division of the limit of the functions, provided that the limit of the denominator is not zero. Thus: \(\lim _ { x \rightarrow b } \frac { f ( x ) } { g ( x ) } = \frac{ \lim _ { x \rightarrow b } f ( x ) }{ \lim _ { x \rightarrow b } g ( x ) } = \frac{7}{-3} = -\frac{7}{3}\).

Key Concepts

Limit of a Sum

When considering the limit of a sum, the principle is straightforward. If you have two functions, say \( f(x) \) and \( g(x) \), and both of their limits as \( x \) approaches some number \( b \) are known, the limit of their sum is simply the sum of their individual limits. This can be symbolically written as: \( \lim_{x \to b} (f(x) + g(x)) = \lim_{x \to b} f(x) + \lim_{x \to b} g(x) \). For our given problem, these limits were 7 and -3, respectively, leading to the sum:

• 7 + (-3) = 4

This rule allows us to effortlessly find the limit of complex expressions by breaking them into simpler parts.

Limit of a Product

The limit of a product is another fundamental rule in calculus. This method states that the limit of the product of two functions is the product of their limits. Simply put, if you know the limits of \( f(x) \) and \( g(x) \) as \( x \) approaches \( b \), you can multiply these limits to find the limit of their product: \( \lim_{x \to b} (f(x) \cdot g(x)) = \lim_{x \to b} f(x) \cdot \lim_{x \to b} g(x) \). For example, with our given limits of 7 and -3, the calculation goes as follows:

• 7 \cdot (-3) = -21

This concise approach helps in quickly solving limits involving products without separately handling complex function multiplication.

Constant Multiple Rule

In the realm of limits, the constant multiple rule simplifies numerous calculations. This principle posits that the limit of a constant multiplied by a function's limit equals the constant times the limit of the function. Expressed mathematically: \( \lim_{x \to b} c \cdot g(x) = c \cdot \lim_{x \to b} g(x) \), where \( c \) is a constant. When calculating with \( c = 4 \) and the limit of \( g(x) = -3 \), we have:

• 4 \cdot (-3) = -12

This simplification is beneficial in problems where functions are scaled by a constant numerical factor.

Limit of a Quotient

Finally, the limit of a quotient is critical in calculus. The limit of the division of two functions is expressed as the division of their limits, provided the denominator's limit isn't zero: \( \lim_{x \to b} \frac{f(x)}{g(x)} = \frac{\lim_{x \to b} f(x)}{\lim_{x \to b} g(x)} \). It's essential to confirm that the denominator's limit isn't zero to avoid undefined expressions. With our functions \( f(x) \) and \( g(x) \), where their limits are 7 and -3, we compute:

• \( \frac{7}{-3} = -\frac{7}{3} \)

This example showcases efficient application of this rule for smoothly handling intricate quotient expressions.