Problem 66 - Calculus

Question

In Problems \(64-71\), find a value of the constant \(k\) such that the limit exists. $$\lim _{x \rightarrow-2} \frac{x^{2}+4 x+k}{x+2}$$

Step-by-Step Solution

Answer

The value of \( k \) is 4.

Step 1: Analyze the Limit

The problem asks us to find a value for the constant \( k \) such that the limit \( \lim_{x \to -2} \frac{x^2 + 4x + k}{x + 2} \) exists. To determine this, recognize that when \( x = -2 \), the denominator becomes zero, causing an indeterminate form if the numerator is also zero at \( x = -2 \).

Step 2: Set up the Condition for the Limit to Exist

For the limit to exist, the expression \( x^2 + 4x + k \) must be divisible by \( x + 2 \), ensuring that it cancels the \( x + 2 \) in the denominator. This means \( x + 2 \) must be a factor of \( x^2 + 4x + k \).

Step 3: Perform Polynomial Division

Use polynomial division or synthetic division on \( x^2 + 4x + k \) by \( x + 2 \). When dividing, the remainder must be zero for \( x + 2 \) to be a factor.

Step 4: Apply the Remainder Theorem

The Remainder Theorem implies that if \( x + 2 \) is a factor, then substituting \( -2 \) into \( x^2 + 4x + k \) gives zero. Substitute \( x = -2 \) and solve: \(((-2)^2) + 4(-2) + k = 0 \Rightarrow 4 - 8 + k = 0 \).

Step 5: Solve for \(k\)

Simplify the expression: \(-4 + k = 0\). Solving this gives \( k = 4 \).

Key Concepts

Polynomial Division

Imagine that you have a polynomial and you want to divide it by another polynomial. This is much like long division with numbers, but a bit more complex due to variables. Polynomial division allows you to determine whether one polynomial is a factor of another. Here's why it's important:

• Helps find roots of polynomials.
• Tests divisibility, answering if one polynomial divides evenly into another with zero remainder.
• Assists in simplifying expressions in calculus to calculate limits.

To divide a polynomial by another, set up the division as you would with numerical long division. Align terms by powers of the variable. Basically, you'll:

• Divide the leading term of the dividend by the leading term of the divisor.
• Multiply the entire divisor by this quotient.
• Subtract the result from the dividend.
• Repeat until the remainder's degree is less than the divisor's.

The process goes on until you either reach a remainder (indicating the divisor is not a factor) or you get zero (indicating it is a factor). Polynomial division is crucial for solving the exercise because it helps us test if \( (x + 2) \) is a factor of \( x^2 + 4x + k \).

Remainder Theorem

The Remainder Theorem is a fascinating tool. It simplifies the process of finding remainders for polynomial division. According to it, if you divide a polynomial \( f(x) \) by \( x - r \), the remainder of that division is \( f(r) \). This means if \( f(r) = 0 \), then \( x - r \) is a factor of \( f(x) \).
In our scenario, we apply the Remainder Theorem by substituting \( x = -2 \) into the polynomial \( x^2 + 4x + k \). The condition for \( x + 2 \) to be a factor is that the remainder should be zero. Thus, \( (-2)^2 + 4(-2) + k = 0 \). This equation must be solved for \( k \) to confirm the existence of a factor, and therefore, the limit.
The beauty of this theorem lies in its simplicity, avoiding complex algebraic operations when testing potential factors of polynomials. It's an essential shortcut that gives quick insights into polynomial behavior.

Indeterminate Forms

In calculus, indeterminate forms are expressions that do not initially provide enough information to evaluate a limit. They come in various types, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms prompt further simplification to calculate the limit's actual value. Common strategies include algebraic manipulation, factoring, or using calculus techniques like L'Hôpital's Rule.

• In the exercise, the expression \( \lim_{x \rightarrow -2} \frac{x^2 + 4x + k}{x+2} \) initially seems to be an indeterminate form \( \frac{0}{0} \).
• This happens because substituting \( x = -2 \) results in both the numerator and denominator becoming zero at the same time.
• To resolve the indeterminate nature, we checked if \( x + 2 \) is a factor of the numerator.

Once a factor cancels out, the problem becomes tractable. Removing indeterminacy helps compute the limit accurately, showing that notions like these are more than just hurdles—they guide us to deeper understanding of calculus and its functionalities.