Problem 28 - Thomas Calculus

Question

Limits of quotients Find the limits in Exercises \(23-42\) $$\lim _{t \rightarrow-1} \frac{t^{2}+3 t+2}{t^{2}-t-2}$$

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{3}\).

Step 1: Identify limit type

First, we need to check if directly substituting the value of t leads to zero in either the numerator or the denominator. Substitute \(t = -1\) into both numerator and denominator.

Step 2: Substitution in the numerator

Substitute \(t = -1\) into \(t^2 + 3t + 2\), which gives \((-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0\). The numerator equals zero.

Step 3: Substitution in the denominator

Substitute \(t = -1\) into \(t^2 - t - 2\), which gives \((-1)^2 - (-1) - 2 = 1 + 1 - 2 = 0\). The denominator also equals zero.

Step 4: Factor both numerator and denominator

Since both parts give zero, we factor both the numerator \((t^2 + 3t + 2)\) and the denominator \((t^2 - t - 2)\). First, factor the numerator: \(t^2 + 3t + 2 = (t + 1)(t + 2)\). Next, factor the denominator: \(t^2 - t - 2 = (t + 1)(t - 2)\).

Step 5: Simplify the expression

We can now cancel the common factor \((t + 1)\) from the numerator and denominator, as long as \(t eq -1\). The expression becomes: \(\frac{(t + 2)}{(t - 2)}\).

Step 6: Substitute again and find the limit

Now substitute \(t = -1\) back into the simplified expression: \(\frac{-1 + 2}{-1 - 2} = \frac{1}{-3} = -\frac{1}{3}\).

Key Concepts

Factoring Polynomials

Factoring polynomials is a vital skill in algebra and calculus. It involves breaking down a polynomial into simpler components, called factors, that when multiplied together give you the original polynomial. In our problem, we are dealing with the expression \(t^2 + 3t + 2\) and \(t^2 - t - 2\). The goal is to express each polynomial as a product of its factors.

• For \(t^2 + 3t + 2\), we find that it can be factored as \((t + 1)(t + 2)\), meaning these are the values of \(t\) that will make the expression zero.
• For \(t^2 - t - 2\), we factor it as \((t + 1)(t - 2)\).

Factoring helps us simplify algebraic expressions, solve equations, and in this case, identify and cancel common factors within rational expressions, making them easier to evaluate under certain operations such as limits.

Indeterminate Forms

Indeterminate forms arise in calculus primarily when evaluating limits. They occur when direct substitution in a limit initially results in expressions like \(\frac{0}{0}\). Such forms do not convey enough information about the limit's behavior, and alternative methods must be employed to find it.
In our exercise, substituting \(t = -1\) in both the numerator and the denominator gives zero, creating the indeterminate form \(\frac{0}{0}\).

• This tells us the original limit cannot be resolved by simple substitution and indicates the potential need for techniques like factoring to simplify the expression first.
• Once simplified, the problem can typically be evaluated by substituting the variable again.

Understanding indeterminate forms is crucial as they are common in limit problems, and knowing how to resolve them is key to successfully solving these types of problems.

Limit Computation

Computing limits is a cornerstone of calculus, providing insight into behaviors of functions as inputs approach certain values. After recognizing an indeterminate form, we use algebraic manipulation to clarify the function's behavior.
In our exercise, after factoring and simplifying \(\frac{t^2 + 3t + 2}{t^2 - t - 2}\) to \(\frac{t + 2}{t - 2}\), we substitute \(t = -1\) in this new, simplified expression. The limit becomes \(-\frac{1}{3}\). This process involves:

• Simplifying the expression by canceling out common factors, reducing the complexity of the limit.
• Re-evaluating the limit using the simplified rational function.
• Understanding that this new result accurately represents the behavior of the original function near the point of interest.

Limit computation often employs additional techniques like L'Hôpital's Rule or using known limits, but in cases such as ours, algebraic simplification often suffices. This skill is essential for understanding fundamental concepts of continuity and calculus applications.