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 \(\infty\) : $$ \lim _{x \rightarrow \infty}\left(1+\frac{1}{x^{p}}\right)^{x} $$

Step-by-Step Solution

Answer

1 when \( p > 1 \); \( e \) when \( p = 1 \); \( \infty \) when \( p < 1 \).

Step 1: Understanding the Limit

The problem asks us to evaluate the limit \( \lim_{x \rightarrow \infty}\left(1+\frac{1}{x^{p}}\right)^{x} \) for different values of \( p > 0 \). We need to determine when this limit equals 1, \( \infty \) or a constant different from 1 and \( \infty \).

Step 2: Simplifying the Internal Expression

Rewrite the expression \( \left(1+\frac{1}{x^{p}}\right)^{x} \) as \( e^{x \ln\left(1+\frac{1}{x^{p}}\right)} \). This helps us analyze the behavior of the limit as \( x \) approaches \( \infty \).

Step 3: Expanding the Logarithm

Use the logarithmic expansion \( \ln(1+u) \approx u \) for \( u \rightarrow 0 \). Substitute \( u = \frac{1}{x^p} \), hence \( \ln\left(1+\frac{1}{x^{p}}\right) \approx \frac{1}{x^p} \).

Step 4: Evaluating the Exponent

Now substitute the approximation back: \( x \ln\left(1+\frac{1}{x^{p}}\right) \approx x \cdot \frac{1}{x^p} = \frac{x}{x^p} = x^{1-p} \).

Step 5: Analyzing Different Cases for \(p\)

- **Case 1:** If \( 1 - p < 0 \) (i.e., \( p > 1 \)), then \( x^{1-p} \rightarrow 0 \), so \( e^{x^{1-p}} \rightarrow e^0 = 1 \).- **Case 2:** If \( 1 - p = 0 \) (i.e., \( p = 1 \)), then \( x^{1-p} = 1 \) which implies \( e^{x^{1-p}} = e^1 = e \). A constant different from 1 and \( \infty \).- **Case 3:** If \( 1 - p > 0 \) (i.e., \( p < 1 \)), then \( x^{1-p} \rightarrow \infty \), so \( e^{x^{1-p}} \rightarrow \infty \).

Step 6: Conclusion

Summarize the findings:- For \( p > 1 \), the limit is 1.- For \( p = 1 \), the limit is a constant equal to \( e \) (which is neither 1 nor \( \infty \)).- For \( p < 1 \), the limit is \( \infty \).

Key Concepts

Exponential Growth

Exponential growth describes a process where a quantity increases rapidly under certain conditions. It is a key concept in calculus and is especially significant when evaluating limits that approach infinity. In the given exercise, the expression \( \left(1+\frac{1}{x^{p}}\right)^{x} \) is an example of exponential growth as \( x \) becomes very large.

When a function grows exponentially, small changes in the variable lead to large changes in the function's output. This is why understanding the internal behavior of the expression is crucial. By transforming it using logarithms, we can investigate how this growth unfolds. By expressing the function as \( e^{x \ln\left(1+\frac{1}{x^{p}}\right)} \), it becomes clear how the exponential component \( e \) influences how rapidly the function increases as \( x \) increases.

• If \( p > 1 \), the function stabilizes, leading to a limit of 1 as the exponential influence is neutralized by the power of \( x^{1-p} \).
• When \( p = 1 \), we observe an exponential growth to a constant value \( e \), showing that certain parameter settings can lead to a non-infinite, non-trivial outcome.
• For \( p < 1 \), the exponential growth is unbounded, meaning the function approaches infinity.

Logarithmic Approximation

Logarithmic approximation is an essential tool when analyzing complex expressions. It simplifies terms that are difficult to handle directly, particularly when they approach zero. In the exercise, we make use of the well-known approximation \( \ln(1+u) \approx u \) when \( u \to 0 \). By setting \( u = \frac{1}{x^p} \), we simplify the analysis of the limit.

This approximation converts the logarithmic term into a linear one, making it easier to manage:

• Each step in a limit problem often involves simplifying problematic expressions; here, the natural logarithm simplifies into \( \frac{1}{x^p} \).
• By reducing complex logarithmic forms to linear approximations, we can better predict the asymptotic behavior as variables grow larger.

As \( x \) grows, \( \frac{1}{x^p} \) approaches zero more quickly for larger \( p \). This simplification underscores how logarithms offer a bridge between multiplicative and additive analysis in calculus.

Asymptotic Behavior

Asymptotic behavior in calculus is about understanding how functions behave as they approach a limit, either towards zero, a constant, or infinity. This concept helps us understand what happens to a function as the input grows extremely large. In this exercise, the asymptotic behavior is captured by analyzing \( e^{x^{1-p}} \), which determines the limit outcome based on different values of \( p \).

Asymptotically, the function behaves differently depending on \( p \):

• When \( p > 1 \), \( x^{1-p} \to 0 \) causing the exponent in \( e^{x^{1-p}} \) to diminish, resulting in a limit of 1.
• For \( p = 1 \), \( x^{1-p} \) equals 1, meaning the function\'s output is a stable, constant \( e \).
• If \( p < 1 \), \( x^{1-p} \to \infty \), leading the exponential term \( e^{x^{1-p}} \) to diverge towards infinity.

This type of analysis helps identify the dominant factors affecting the limit. It allows us to discern the prevailing trend when the input trends towards infinite values. In summary, understanding asymptotic behavior is crucial for evaluating limits and distinguishing between finite and infinite potential outputs.