Problem 24 - Calculus and its applications

Question

The initial substitution of \(x=a\) yields the form \(0 / 0 .\) Look for ways to simplify the function algebraically, or use a table or graph to determine the limit. When necessary, state that the limit does not exist. $$ \lim _{x \rightarrow-3} \frac{2 x^{2}-x-21}{9-x^{2}} $$

Step-by-Step Solution

Answer

The limit does not exist as the function approaches a vertical asymptote at \(x = -3\).

Step 1: Check Initial Substitution

Substitute \(x = -3\) into the function to see if it gives the form \(\frac{0}{0}\).\[ f(x) = \frac{2x^2 - x - 21}{9 - x^2} \]Substituting \(x = -3\):\[ \frac{2(-3)^2 - (-3) - 21}{9 - (-3)^2} = \frac{18 + 3 - 21}{9 - 9} = \frac{0}{0}\] Since the result is \(\frac{0}{0}\), further simplification is needed.

Step 2: Factor the Numerator and Denominator

Attempt to factor both the numerator \(2x^2 - x - 21\) and the denominator \(9 - x^2\).For the numerator:\[ 2x^2 - x - 21 = (2x + 7)(x - 3) \]For the denominator:\[ 9 - x^2 = (3 - x)(3 + x) \]

Step 3: Simplify the Expression

Now substitute the factored forms into the original expression to simplify.\[ \frac{(2x + 7)(x - 3)}{(3 - x)(3 + x)} \]Notice that \((x - 3)\) can be rewritten using \((3-x)\):\[ x - 3 = -(3 - x) \]Substitute this back:\[ \frac{(2x + 7)(-(3 - x))}{(3 - x)(3 + x)} = -\frac{(2x + 7)}{(3 + x)}, x eq 3 \]

Step 4: Evaluate the Simplified Limit

Evaluate the simplified expression \(-\frac{(2x + 7)}{(3 + x)}\) as \(x\) approaches \(-3\).Substitute \(x = -3\):\[ -\frac{(2(-3) + 7)}{(3 - 3)} = -\frac{-6 + 7}{0} = -\frac{1}{0} \]This shows that there is a vertical asymptote and the limit approaches infinity.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are expressions that do not have a definitive limit or straightforward solution at first glance. A common indeterminate form is \(\frac{0}{0}\), which often arises when you directly substitute a value into a limit expression. This occurs because both the numerator and denominator independently approach zero, making it unclear what the limit of the function should be.The exercise we are examining initially presents this \(\frac{0}{0}\) form when substituting \(x = -3\) into the original function:

• The numerator, \(2x^2 - x - 21\), and the denominator, \(9 - x^2\), both equal zero at \(x = -3\), leading to the indeterminate form.

To resolve an indeterminate form, algebraic manipulation like factoring can simplify the expression and help find the limit. This eliminates the zero terms and reveals the function's behavior as \(x\) approaches the given value.

Factoring Polynomials

Factoring is a key technique used in calculus to simplify polynomials. It involves breaking down a complex expression into the product of simpler factors, making it easier to simplify or evaluate as part of solving limits.In the provided solution, both the numerator and the denominator are factored to assist in simplifying the limit:

• The quadratic polynomial in the numerator, \(2x^2 - x - 21\), factors into \((2x + 7)(x - 3)\).
• The difference of squares in the denominator, \(9 - x^2\), factors into \((3 - x)(3 + x)\).

These factored forms allow for the cancellation of common factors—particularly important when one factor of both the numerator and the denominator creates the indeterminate form (like \(x - 3\)). Once these common factors are cancelled, the simplified expression can more easily be evaluated for the limit.

Vertical Asymptotes

Vertical asymptotes occur in functions where the limit approaches infinity or negative infinity as \(x\) approaches a certain value. This is often the result when the denominator of a function tends toward zero while the numerator approaches a non-zero constant.In the exercise solution, after simplifying, the expression becomes:\[-\frac{(2x + 7)}{(3 + x)}\]As \(x\) approaches \(-3\), the denominator \(3 + x\) approaches zero, but the numerator converges to \(-1\.\)

• This indicates there is a vertical asymptote at \(x = -3\). As \(x\) gets closer to \(-3\), the function value increasingly tends towards infinity or negative infinity.

Understanding vertical asymptotes helps provide insights into the behavior of functions near certain values where division by zero would otherwise be apparent.