Problem 108 - Thomas Calculus

Question

$$ \lim _{n \rightarrow \infty} x^{1 / n}=1,(x>0) $$

Step-by-Step Solution

Answer

The limit of \(x^{1/n}\) as \(n\to \infty\) is 1 if \(x>0\).

Step 1: Understand the Exercise

The exercise requires us to prove that the limit of the expression \( x^{1/n} \) as \( n \) approaches infinity equals 1, given that \( x > 0 \). This involves understanding the behavior of the expression as \( n \) becomes very large.

Step 2: Consider the Definition of Root

The expression \( x^{1/n} \) can be rewritten as the \( n \)-th root of \( x \). As \( n \) approaches infinity, the \( n \)-th root of \( x \) approaches 1 because each additional degree in the root makes the expression tend more and more toward flattening at 1.

Step 3: Use Logarithms to Simplify

Take the natural logarithm of both sides: \( \ln(y) = \ln(x^{1/n}) = \frac{1}{n}\ln(x) \). This simplifies the expression using the power rule of logarithms.

Step 4: Evaluate the Limit of the Logarithm

Evaluate the limit \( \lim_{n \to \infty} \frac{1}{n} \ln(x) \). Since \( \ln(x) \) is constant and \( \frac{1}{n} \) approaches 0 as \( n \to \infty \), the limit of \( \frac{1}{n} \ln(x) \) is 0.

Step 5: Exponentiate the Result

Now exponentiate both sides: \( y = e^0 = 1 \). This concludes that \( \lim_{n \to \infty} x^{1/n} = 1 \).

Key Concepts

n-th Root

The concept of the n-th root is essential for understanding the behavior of expressions like \( x^{1/n} \). When you see \( x^{1/n} \), it's useful to think about it as the n-th root of \( x \). As \( n \) gets larger, the n-th root has an interesting property: it tends to move the value closer to 1 for any positive \( x \).

Why does this happen? Consider the n-th root as a kind of 'flattening' process. Each added degree of the root causes the value to decrease faster toward 1. Imagine large numbers: if you repeatedly apply the root operation, you diminish the original value's influence, causing the entire expression to converge to a steady point. In the limit as \( n \) approaches infinity, this point is exactly 1. This phenomenon occurs for any positive number \( x \), demonstrating a fundamental feature of the n-th root.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), helps simplify the complexity of exponential expressions. Taking the natural logarithm of both sides of \( x^{1/n} \) allows us to use logarithmic properties to make the problem more manageable.

Using the power rule of logarithms, \( \ln(x^{1/n}) \) can be rewritten as \( \frac{1}{n} \ln(x) \). This transformation is powerful because it reduces an exponential expression to a linear one. You no longer deal with the exponent directly, but instead with a product, which is often much easier to analyze.

In the context of limits, this simplification is particularly useful. When you take the limit \( \lim_{n \to \infty} \frac{1}{n} \ln(x) \), the term \( \frac{1}{n} \) becomes very small as \( n \) increases, turning the entire expression toward zero. Hence, applying the natural logarithm has made evaluating this limit much more straightforward.

Exponential Function

The exponential function, denoted as \( e^x \), functions as the 'undo' operation when using natural logarithms in our calculations. After finding that \( \lim_{n \to \infty} \frac{1}{n} \ln(x) = 0 \), we use the exponential function to revert back.

The exponential function and natural logarithms are inverse operations. Thus, when we had \( \ln(y) = 0 \), exponentiating both sides with \( e \) gives \( y = e^0 \). Since \( e^0 = 1 \), this result completes the process started with the logarithm, showing that indeed \( \lim_{n \to \infty} x^{1/n} = 1 \).

Not only does the exponential function serve as a reversal tool, but it also confirms the limit result by converting the logarithmic solution back to an exponential form. This interplay between logarithms and exponentials is a cornerstone of calculus, allowing us to simplify and solve complex problems with limits.