Problem 15 - Calculus: Early Transcendentals

Question

Evaluate the limit, if it exists. \( \displaystyle \lim_{t \to -3}\frac{t^2 - 9}{2t^2 + 7t + 3} \)

Step-by-Step Solution

Answer

The limit exists and is \( \frac{6}{5} \).

Step 1: Plug in the value

First, try substituting the value of the limit directly into the expression. Substitute \( t = -3 \) into \( \frac{t^2 - 9}{2t^2 + 7t + 3} \). \[ \frac{(-3)^2 - 9}{2(-3)^2 + 7(-3) + 3} = \frac{9 - 9}{18 - 21 + 3} = \frac{0}{0} \] The result is an indeterminate form \( \frac{0}{0} \), so we need further simplification.

Step 2: Factor the numerator and the denominator

Factor both the numerator and the denominator to simplify.The numerator \( t^2 - 9 \) is a difference of squares, which factors to \((t - 3)(t + 3)\).The denominator \( 2t^2 + 7t + 3 \) can be factored by grouping:- Find two numbers that multiply to \( 2 \times 3 = 6 \) and add to \( 7 \), which are \( 6 \) and \( 1 \).- Rewrite and group: \( 2t^2 + 6t + t + 3 = (2t^2 + 6t) + (t + 3) \).- Factor each group: \( 2t(t + 3) + 1(t + 3) \).- Factor out the common factor: \( (t + 3)(2t + 1) \).Now write the expression:\[ \frac{(t - 3)(t + 3)}{(t + 3)(2t + 1)}\]

Step 3: Simplify the expression

Cancel the common factor \((t + 3)\) from both the numerator and the denominator:\[ \frac{t - 3}{2t + 1} \]This step is allowed as long as \( t eq -3 \).

Step 4: Calculate the limit

Now, substitute \( t = -3 \) directly into the simplified expression:\[ \lim_{t \to -3} \frac{t - 3}{2t + 1} = \lim_{t \to -3} \frac{-3 - 3}{2(-3) + 1} = \frac{-6}{-6-1} = \frac{-6}{-6+1} = \frac{-6}{-5} = \frac{6}{5} \]

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms often occur when evaluating limits. An expression is classified as an indeterminate form when direct substitution of the limit value leads to an undefined operation, usually appearing as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 0 \times \infty \), among others. In our exercise, substituting \( t = -3 \) into the function \( \frac{t^2 - 9}{2t^2 + 7t + 3} \) initially results in \( \frac{0}{0} \).

• When you encounter an indeterminate form, it signifies that further algebraic manipulation or calculus techniques are necessary to find the limit.
• Indeterminate forms suggest that the surface-level substitution does not provide insightful information about the behavior of the function around the point of interest.
• In our example, the solution involves factoring the polynomial expressions to simplify and evaluate the limit correctly.

Factoring Polynomials

Factoring polynomials is a key technique used in calculus to simplify expressions, particularly for evaluating limits involving rational functions. In our given problem, both the numerator \( t^2 - 9 \) and the denominator \( 2t^2 + 7t + 3 \) can be factored to help overcome the initial indeterminate form.

• Factoring is transforming a complicated polynomial into a product of simpler polynomials, making it easier to analyze or simplify.
• The numerator is recognized as a difference of squares: \( t^2 - 9 = (t - 3)(t + 3) \).
• For the denominator, factor by grouping is applied, resulting in \( (t + 3)(2t + 1) \).

Using factoring, the expression simplifies and allows cancellation of terms that create indeterminacy.

Difference of Squares

The difference of squares is a particular type of algebraic expression that can be factored into a specific form. A difference of squares is expressed as \( a^2 - b^2 \), where it can always be factored to: \( (a - b)(a + b) \).

• This pattern emerges frequently in calculus, especially when simplifying polynomials during limit evaluations.
• In our exercise, the numerator \( t^2 - 9 \) fits this pattern since \( t^2 \) and \( 9 \) are both perfect squares, making it \( (t - 3)(t + 3) \).

Identifying and using the difference of squares allows simplification of complex expressions and is an essential skill when handling polynomial limits.

Simplifying Expressions

Simplification is the process of making an algebraic expression as straightforward as possible. After factoring, as demonstrated in our exercise, simplifying involves canceling out common factors and clarifying the expression.

• In our problem, after the factorization of the numerator and denominator, we notice that \( (t + 3) \) is a common factor.
• Cancelling \( (t + 3) \) yields \( \frac{t - 3}{2t + 1} \), provided \( t eq -3 \).
• Finally, evaluate the simplified expression by substituting \( t = -3 \) resulting in the limit \( \frac{6}{5} \).

Through simplification, we're able to resolve the initial indeterminate form and determine the true behavior of the function near the limit point.