Problem 63 - Calculus for Biology and Medicine

Question

For \(p>0\), determine the values of \(p\) for which the following limit is either 1 or \(\infty\) or a constant that is neither 1 nor \(\infty\) : $$ \lim _{x \rightarrow \infty}\left(1+\frac{c}{x^{p}}\right)^{x} $$

Step-by-Step Solution

Answer

For \( p = 1 \), the limit is \( e^c \); for \( p < 1 \), it is \( \infty \); for \( p > 1 \), it is 1.

Step 1: Identify the Limit Form

We need to evaluate the limit \( \lim _{x \rightarrow \infty}\left(1+\frac{c}{x^{p}}\right)^{x} \). This expression is of the form \( \left(1 + \text{small quantity} \right)^{\text{large quantity}} \), which resembles the exponential limit form.

Step 2: Exponential Limit Transformation

Recall that \( \left(1 + \frac{1}{n}\right)^n \to e \) as \( n \to \infty \). Therefore, let's set \( a = \frac{c}{x^p} \). As \( x \to \infty \), \( a \to 0 \). The limit can thus be transformed by taking the natural logarithm: \[ \lim_{x \to \infty} \left( 1 + \frac{c}{x^p} \right)^{x} = \exp \left( \lim_{x \to \infty} x \ln \left( 1 + \frac{c}{x^p} \right) \right) \]

Step 3: Simplify Using Taylor Series

For small \( a \), we use the approximation \( \ln(1+a) \approx a \). Therefore, \( \ln \left( 1 + \frac{c}{x^p} \right) \approx \frac{c}{x^p} \).Substitute it back into the limit expression: \[ \lim_{x \to \infty} x \cdot \frac{c}{x^p} = c \cdot \lim_{x \to \infty} \frac{x}{x^p} = c \cdot \lim_{x \to \infty} x^{1-p} \]

Step 4: Determine the Behavior based on p

Evaluate the limit \( \lim_{x \to \infty} x^{1-p} \) depending on the value of \( p \):- If \( p = 1 \), the limit becomes \( c \cdot \lim_{x \to \infty} x^{0} = c \), a constant, since \( x^0 = 1 \).- If \( p < 1 \), then \( x^{1-p} \to \infty \) as \( x \to \infty \), making the exponential limit \( \infty \).- If \( p > 1 \), then \( x^{1-p} \to 0 \) as \( x \to \infty \), so the exponential limit converges to \( e^0 = 1 \).

Key Concepts

Exponential Limit

The concept of an exponential limit involves expressions that have the structure \( ig(1 + ext{small quantity}\big)^{ ext{large quantity}} \). This is a hallmark of exponential behavior in calculus. When we see this form, we are reminded of the fundamental limit result: \( \left(1 + \frac{1}{n}\right)^n \to e \) as \( n \to \infty \). This principle helps us evaluate limits that seem complex at first glance.
By transforming the expression into a format that exploits the exponential limit property, we can simplify and solve them. In many cases, taking the natural logarithm can reveal this structure more clearly and make the expression manageable to evaluate. They are especially useful when dealing with terms that become infinitesimally small as another quantity grows indefinitely large. This transformation from a potentially complicated expression into a more familiar form is one of the cornerstones of understanding limits in calculus.

Taylor Series

The Taylor Series is an essential tool used to approximate functions that can contact small terms, allowing us to simplify complex calculations. When dealing with limits, especially those involving logarithmic functions, it can provide a convenient approximation.
For expressions like \( \ln(1 + a) \) when \( a \) is small, we often use the fact that \( \ln(1 + a) \approx a \). This simplification stems from the Taylor series expansion of \( \ln(1 + a) \), which is: \( a - \frac{a^2}{2} + \frac{a^3}{3} - \cdots \).
By using only the first term of this expansion, we are making a practical approximation that captures the essence of the function's behavior for small arguments. This is particularly useful when examining limits where the variable approaches infinity, as it streamlines the equation to something much more functionally approachable in terms of its primary contributing factor.

Behavior of Limits

Understanding the behavior of limits as variables approach certain thresholds, like infinity, is crucial in calculus. The problem given is resolved by considering how the limit behaves based on the parameter \( p \).
- When \( p = 1 \), the term \( x^{1-p} \) simplifies to \( x^0 = 1 \), and thus the limit is constant, \( c \).
- If \( p < 1 \), then \( x^{1-p} \) grows without bound, leading the whole expression to approach infinity.
- Conversely, if \( p > 1 \), \( x^{1-p} \) diminishes towards zero, making the limit simplify to \( e^0 = 1 \).
These scenarios highlight the importance of analyzing expressions to predict their convergent or divergent nature as parameters change. Such assessments allow us to interpret quite effectively what an expression tends towards, offering valuable insights into its long-term behavior.