Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Show that if \(B\) is a number greater than \(1,\) and \(n\) is a positive integer, $$ \lim _{t \rightarrow \infty} \frac{B^{t}}{t^{n}}=\infty $$

Step-by-Step Solution

Answer

\( \frac{B^t}{t^n} \) tends to infinity as \( t \to \infty \) because \( B^t \) grows exponentially.

Step 1: Define the Problem

We need to show that the limit \( \lim _{t \rightarrow \infty} \frac{B^{t}}{t^{n}}=\infty \) holds true for a fixed real number \( B > 1 \) and a positive integer \( n \). In simple terms, as \( t \) becomes larger and larger, \( \frac{B^t}{t^n} \) becomes infinitely large.

Step 2: Understand Exponential Growth

Recognize that \( B^t \) is an exponential function with base \( B > 1 \). This means that as \( t \) increases, \( B^t \) grows exponentially and becomes very large quite rapidly.

Step 3: Compare Growth Rates

Forward the concept that exponential functions grow faster than polynomial functions. Specifically, for sufficiently large \( t \), the dominance of the exponential term \( B^t \) over the polynomial term \( t^n \) will cause the entire ratio \( \frac{B^t}{t^n} \) to increase towards infinity.

Step 4: Use L'Hôpital's Rule (Optional)

To rigorously justify the limit, consider using L'Hôpital's Rule. While this rule applies for limits of the form \( \frac{f(t)}{g(t)} \to \frac{\infty}{\infty} \), you would repeatedly apply it until the polynomial term disappears, highlighting that the numerator \( B^t \) grows faster. However, careful application here may require multiple steps depending on \( n \).

Step 5: Formal Limit Argument

For a direct argument, let's leverage the inequality: for any \( \epsilon > 0 \), there exists some \( t_0 \) such that for all \( t > t_0 \), \( B^t > \epsilon \cdot t^n \). By letting \( \epsilon = 1 \), or smaller, show that \( \frac{B^t}{t^n} > \epsilon \) leads to infinity as \( t \) approaches infinity.

Step 6: Conclusion

Since \( B^t \) grows unboundedly while \( t^n \) grows polynomially, the fraction \( \frac{B^t}{t^n} \) also grows without bound. Therefore, \( \lim _{t \rightarrow \infty} \frac{B^t}{t^n} = \infty \) is verified.

Key Concepts

Exponential Growth

When talking about exponential growth, it's essential to understand what the base and the exponent do. In the expression \( B^t \), \( B \) is the base, and it's consistently greater than one. The exponent \( t \) shows how many times we multiply \( B \) by itself. An exponential function increases rapidly as \( t \) becomes larger.
\( B^t \) grows exponentially because each additional step multiplies the total by \( B \). For instance:

• If \( B = 2 \) and \( t = 3, \) the result is \( 2 \times 2 \times 2 = 8 \).
• If \( t = 4, \) the value becomes \( 2 \times 2 \times 2 \times 2 = 16 \).

From this example, you can see how quickly the numbers grow. As \( t \) continues to increase, the values of \( B^t \) become much larger very fast. This is a hallmark of exponential growth, and it becomes quite essential when comparing it against other types of growth patterns.

Polynomial Functions

Polynomial functions, in contrast to exponential functions, increase more slowly. A polynomial function has the form \( t^n \), where \( n \) is a positive integer known as the degree of the polynomial.
This means that our function is growing in a regular pattern, but it is not as explosive as exponential growth.
For example:

• If \( n = 2 \) and \( t = 3, \) \( t^2 = 3^2 = 9 \).
• If \( t = 4, \) \( t^2 = 4 \times 4 = 16 \).

This growth is called polynomial because it keeps multiplying \( t \) but only a set number of times specified by \( n \). Compared to exponential growth, which can multiply the base infinitely as \( t \) increases, a polynomial will not grow as fast. This slower growth is why in the formula \( \frac{B^t}{t^n} \), the exponential part \( B^t \) eventually overtakes \( t^n \), leading the whole fraction to infinity.

L'Hôpital's Rule

L'Hôpital's Rule is a calculus tool used mostly to find limits that involve indeterminate forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This can be particularly useful when examining the limit \( \lim _{t \rightarrow \infty} \frac{B^t}{t^n} \).
In essence, L'Hôpital's Rule states that under specific conditions, you can take the derivative of the numerator and the denominator until the limit becomes solvable. Consider the function \( f(t) = B^t \) and \( g(t) = t^n \). When \( t \) approaches infinity, both functions will head towards infinity, creating an indeterminate form of \( \frac{\infty}{\infty} \).

• By applying L'Hôpital's Rule, you take the derivatives: \( f'(t) = B^t \ln B \) and \( g'(t) = n t^{n-1} \).
• Repeating this step, the polynomial eventually flattens out, while the exponential term remains strong.

This continuous approach highlights that because \( B^t \) loses none of its exponential nature, and the polynomial fades with derivation, the limit of the fraction tends towards infinity. That's why L'Hôpital's Rule, although optional, provides an assurance that \( \frac{B^t}{t^n} \) becomes infinitely large.