Problem 11 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

In Equation 16.11 , we claim that for $$ x_{t}=C_{1} r_{1}^{t}+C_{2} \times t \times r_{1}^{t}, \quad \lim _{t \rightarrow \infty} A_{t}=0 \quad \text { if } \quad\left|r_{1}\right|<1 $$ Use L'Hospital's Theorem 14.5.1 to show that \(\lim _{t \rightarrow \infty} t \times r^{t}=0\) if \(|r|<1\).

Step-by-Step Solution

Answer

The limit \(\lim_{t \rightarrow \infty} t \times r^t = 0\) if \(|r| < 1\).

Step 1: Express the Limit to Evaluate

First, we need to express the limit we want to evaluate:\[\lim_{t \rightarrow \infty} t \times r^t\]where \(|r| < 1\). This is an indeterminate form of type \(\infty \times 0\).

Step 2: Reformat as a Fraction

Transform the expression into a form suitable for L'Hôpital's rule. Rewrite it as:\[\lim_{t \rightarrow \infty} \frac{t}{r^{-t}}\]This substitution converts it into the \(\frac{\infty}{\infty}\) indeterminate form.

Step 3: Apply L'Hôpital's Rule

Since the limit is in the \(\frac{\infty}{\infty}\) form, apply L'Hôpital's Rule:Differentiating the numerator: \(\frac{d}{dt}[t] = 1\).Differentiating the denominator: \(\frac{d}{dt}[r^{-t}] = -r^{-t} \ln(r)\).Hence, the limit becomes:\[\lim_{t \rightarrow \infty} \frac{1}{-r^{-t} \ln(r)}\]

Step 4: Evaluate the New Limit Expression

Rewrite the limit:\[\lim_{t \rightarrow \infty} \frac{1}{-r^{-t} \ln(r)} = \lim_{t \rightarrow \infty} \frac{1}{-(1/r)^t \ln(r)}\]Because \(|r| < 1\), \(\frac{1}{r} > 1\), so \((1/r)^t \rightarrow \infty\) as \(t \rightarrow \infty\). The denominator thus tends to infinity, making the entire fraction approach zero.

Step 5: Conclude the Limit Evaluation

Hence, using L'Hôpital's Rule, we find:\[\lim_{t \rightarrow \infty} t \times r^t = 0\]which confirms the claim made in the problem.

Key Concepts

Limits

Limits are a fundamental concept in calculus, representing the behavior of a function as it approaches a certain point. We often express this as \ \( \lim_{t \rightarrow a} f(t) = L \ \, \), where \ \(L\ \) is the function's value as \ \(t\ \) approaches \ \(a\ \). In simple terms, a limit helps us understand what happens to a function the closer its input gets to a particular value.

It's an essential tool when dealing with functions that are not easily expressed at all points.

• For example, consider \ \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \ \).
• This indicates that as \ \(x\ \) gets closer and closer to zero, \ \(\frac{\sin x}{x}\ \) approaches the value of 1.

In the context of our problem, limits allow us to analyze the behavior of the sequence as \ \(t\ \) approaches infinity, helping us determine the outcome of expressions that seem undefined at first glance.

Indeterminate Forms

Indeterminate forms arise when evaluating limits that do not initially reveal their behavior. These include forms like \( \frac{0}{0} \), \( \infty - \infty \), or \( \infty \times 0 \), which can usually be tackled with techniques such as algebraic manipulation or calculus tools like L'Hôpital's Rule.

Our exercise deals with the form \( \infty \times 0 \), when evaluating \( \lim_{t \rightarrow \infty} t \times r^t \). Despite each factor tending towards opposite extremes—one to infinity and the other to zero—their combined behavior can sometimes yield a finite result, often zero.

• Here, by transforming the expression \( t \times r^t \) into a fraction, \( \frac{t}{r^{-t}} \, \) we convert it into an \( \frac{\infty}{\infty} \) form.
• It's a manipulation that prepares the problem for L'Hôpital's Rule.

This technique assists us in conquering the ambiguity of such forms, providing a clear answer when the untrained eye might be uncertain.

Calculus Problem Solving

Calculus problem solving often involves navigating through complex expressions and applying the right techniques to find a solution. In this exercise, we use L'Hôpital's Rule, a powerful tool for resolving limits involving indeterminate forms.

Here's how this works in the problem:

• First, recognize an indeterminate form that L'Hôpital's Rule can help resolve, such as \( \frac{\infty}{\infty} \).
• Then, differentiate both the numerator and the denominator separately.
• This often simplifies the limit, making it easier to evaluate.

In the problem at hand, once we've rewritten the original expression \( \lim_{t \rightarrow \infty} \frac{t}{r^{-t}} \), we apply L'Hôpital's Rule:

• The derivative of \( t \, \) the numerator, is 1.
• For the denominator \( r^{-t} \, \) the derivative is \( -r^{-t} \ln(r) \).

Thus, after applying these derivatives in the limit expression, it becomes manageable and leads to a conclusive result that demonstrates the behavior of the function as \( t \, \) approaches infinity.