Problem 15 - Calculus Early Transcendentals

Question

Evaluate the limit, if it exists. $$\lim _{t \rightarrow-3} \frac{t^{2}-9}{2 t^{2}+7 t+3}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Form of the Limit

Evaluate the function at \(t = -3\):\[\frac{(-3)^2 - 9}{2(-3)^2 + 7(-3) + 3} = \frac{9 - 9}{2(9) + (-21) + 3} = \frac{0}{0}\]Since the expression results in a \(\frac{0}{0}\) indeterminate form, we need to simplify the expression further.

Step 2: Factor Numerator and Denominator

Factor the numerator and the denominator:- Numerator: \(t^2 - 9 = (t - 3)(t + 3)\)- Denominator: Factorize \(2t^2 + 7t + 3\). Using the quadratic formula or simple factoring:Set \(2t^2 + 7t + 3 = (2t + 1)(t + 3)\).

Step 3: Simplify the Expression

Cancel common factors in the numerator and denominator:\[\frac{(t - 3)(t + 3)}{(2t + 1)(t + 3)}\]Cancel \((t + 3)\) from both the numerator and the denominator:\[\frac{t - 3}{2t + 1}\]

Step 4: Evaluate the Limit With Simplified Expression

Substitute \(t = -3\) back into the simplified function:\[\frac{-3 - 3}{2(-3) + 1} = \frac{-6}{-6 - 1} = \frac{-6}{-5} = \frac{6}{5}\]Therefore, the limit is \(\frac{6}{5}\).

Key Concepts

Indeterminate Form

When evaluating limits in calculus, we sometimes encounter expressions that are not straightforward to solve. These are called indeterminate forms. An indeterminate form arises when evaluating a limit in situations such as \(\frac{0}{0}\), \(-\infty/\infty\), or \(0/\infty\). These forms do not immediately reveal the limit's value and require further mathematical manipulation to resolve.

For example, consider the original exercise: \\[\lim _{t \rightarrow-3} \frac{t^{2}-9}{2 t^{2}+7 t+3}\]
Evaluating directly by substituting \(t = -3\) results in a \\[\frac{0}{0}\]indeterminate form. This tells us the expression needs to be further simplified to unveil the actual limit value. Recognizing such patterns guides us in deploying techniques like factoring, simplifying, or applying L'Hôpital's rule.

Factoring Polynomials

Factoring is a critical math technique used to simplify polynomial expressions by breaking them down into simpler components, or factors. This makes them easier to manage and manipulate, especially when dealing with limits or integrals.

• To factor a polynomial, identify expressions like \(a^2 - b^2\), which can be split into two binomials: \\[(a - b)(a + b).\]
• Another common approach involves grouping terms or using the quadratic formula for a general quadratic expression \(ax^2 + bx + c\).

In our exercise, the polynomial in the numerator \(t^2 - 9\) factors into \\[(t - 3)(t + 3).\]
Similarly, the denominator \(2t^2 + 7t + 3\) can be expressed as \\[(2t + 1)(t + 3).\]
Such factoring highlights common terms in the fraction, leading us towards simplification.

Simplifying Rational Expressions

After factoring, the next step is simplifying rational expressions. This process involves reducing fractions by canceling out common factors from the numerator and denominator. Simplification is an essential skill in solving limits, making calculations much more manageable.

Consider the factored expression from the exercise:\\[\frac{(t - 3)(t + 3)}{(2t + 1)(t + 3)}\]
The term \(t + 3\) appears in both the numerator and denominator, allowing us to cancel it out:\\[\frac{t - 3}{2t + 1}\]
Now, the limit is no longer an indeterminate form and is ready for evaluation.

Substitution Method

After simplifying the expressions, the final step is evaluating the limit using the substitution method. This involves plugging the point of interest back into the simplified expression to find the limit.

• In problems involving limits, once the rational expression is simplified, substitute the variable's specified value directly into the expression.
• This method works effectively when the indeterminate form is resolved.

In our example, after simplifying, the function becomes:\\[\frac{t - 3}{2t + 1}\]
Substitute \(t = -3\):\\[\frac{-3 - 3}{2(-3) + 1} = \frac{-6}{-6 - 1} = \frac{-6}{-5} = \frac{6}{5}.\]
With substitution, we determine that the limit is \(\frac{6}{5}\), resolving the problem.