Problem 26 - Calculus Early Transcendentals

Question

Determining limits analytically Determine the following limits or state that they do not exist. $$\lim _{t \rightarrow 5} \frac{4 t^{2}-100}{t-5}$$

Step-by-Step Solution

Answer

Question: Determine the limit of the given function analytically as t approaches 5: \(\lim _{t \rightarrow 5} \frac{4 t^{2}-100}{t-5}\). Answer: The limit of the given function as t approaches 5 is 40.

Step 1: Identify the indeterminate form of the given function at t=5

At t=5, the function takes the form \(\frac{0}{0}\) which is an indeterminate form.

Step 2: Factor and simplify the expression

We can factor the numerator: $$4t^2 - 100 = 4(t^2 - 25) = 4(t-5)(t+5)$$ Therefore, the expression can be simplified: $$\frac{4 t^{2}-100}{t-5} = \frac{4(t-5)(t+5)}{t-5}$$ Now, we can cancel out the common factor \((t-5)\) on both numerator and denominator: $$\frac{4(t-5)(t+5)}{t-5} = 4(t+5)$$

Step 3: Find the limit of the simplified expression

Now we can find the limit of the simplified expression as t approaches 5: $$\lim _{t \rightarrow 5} 4(t+5) = 4(5 + 5) = 4(10) = 40$$

Step 4: State the result of the limit

The limit of the given function as t approaches 5 is 40: $$\lim _{t \rightarrow 5} \frac{4 t^{2}-100}{t-5} = 40$$

Key Concepts

Indeterminate Forms

When we examine the expression \(\lim_{t \rightarrow 5} \frac{4t^2 - 100}{t-5}\), we notice that substituting \(t = 5\) results in \(\frac{0}{0}\). This is called an indeterminate form. Indeterminate forms arise in calculus when direct substitution does not provide a clear answer for the limit. They can take various forms like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), or \(0 \cdot \infty\) among others. Handling indeterminate forms requires special techniques such as factoring, rationalizing, or using L'Hôpital's rule. These methods help us to either redefine or simplify the expression so that it can properly compute the limit. In our case, we will use factoring to resolve the indeterminate form.

Factoring Expressions

Factoring is a powerful tool in algebra and calculus that helps to simplify complex expressions. It involves breaking down an expression into a product of its simpler multipliers. In the given function \(4t^2 - 100\), the first step is to recognize this as a difference of squares, which can be factored as follows:

• First, consider the expression \(4t^2 - 100\)
• Notice it forms a difference of squares: \((2t)^2 - (10)^2\)
• This can be factored to \((2t - 10)(2t + 10)\)

However, a simpler factorization is \(4(t-5)(t+5)\), as \(t^2 - 25\) is a difference of squares \((t-5)(t+5)\). After factoring, the expression becomes easier to work with, allowing us to cancel common terms with the denominator \((t-5)\).

Limit Laws

Limit laws are a set of rules that provide a framework for evaluating limits systematically. These rules allow us to simplify complex limit problems by breaking them into simpler parts. Key limit laws include:

• The Sum/Difference Law: \(\lim_{x \rightarrow a} [f(x) \pm g(x)] = \lim_{x \rightarrow a} f(x) \pm \lim_{x \rightarrow a} g(x)\).
• The Product Law: \(\lim_{x \rightarrow a} [f(x) \cdot g(x)] = \lim_{x \rightarrow a} f(x) \cdot \lim_{x \rightarrow a} g(x)\).
• The Quotient Law, applicable when \(\lim_{x \rightarrow a} g(x) eq 0\): \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}\).

In our example, once we've simplified the expression, applying the limit law becomes straightforward. After canceling out the \((t-5)\) terms, the expression \(4(t+5)\) can be directly evaluated as \(t\) approaches 5: \(4(10) = 40\).

Simplifying Rational Expressions

Simplification is key when dealing with rational expressions, especially when they involve indeterminate forms. Simplifying involves reducing the complexity of the expression without changing its value. Here's how it plays out in the example:

• We started with \(\frac{4t^2 - 100}{t-5}\), a rational expression.
• After factoring the numerator as \(4(t-5)(t+5)\), we identified a common factor with the denominator \((t-5)\).
• Cancelling \((t-5)\) from both the numerator and the denominator gave us \(4(t+5)\).

This new, simpler expression not only resolves the indeterminate form but also allows us to evaluate the limit seamlessly. Simplification is crucial for changing the expression into a more manageable form, paving the way for easier limit evaluation.