Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each limit. Hint: Transform to problems involving a continuous variable \(x\). Assume that \(a>0\). (a) \(\lim _{n \rightarrow \infty} \sqrt[n]{a}\) (b) \(\lim _{n \rightarrow \infty} \sqrt[n]{n}\) (c) \(\lim _{n \rightarrow \infty} n(\sqrt[n]{a}-1)\) (d) \(\lim _{n \rightarrow \infty} n(\sqrt[n]{n}-1)\)

Step-by-Step Solution

Answer

(a) 1, (b) 1, (c) \(\ln a\), (d) \(\infty\).

Step 1: Understanding the Hint

The hint suggests transforming the limit expression to involve a continuous variable like \(x\) instead of \(n\). We'll achieve this by using known behaviors of continuous functions.

Step 2: Solve Part (a)

We recognize \(\lim_{n \to \infty} \sqrt[n]{a} = \lim_{n \to \infty} a^{1/n}\). Using the property that for any \(a > 0\), \(a^{1/n} \to 1\) as \(n \to \infty\), the limit is 1.

Step 3: Solve Part (b)

For \(\lim_{n \to \infty} \sqrt[n]{n}\), switch to logarithms and use L'Hôpital's Rule. The limit can be rewritten as \(\lim_{n \to \infty} e^{\frac{\ln n}{n}}\). Differentiating top and bottom: numerator \(\ln n\) and \(1\) gives \(\frac{1}{n}\), and \(n\) and \(1\) gives \(0\). Thus, the limit is \(e^0 = 1\).

Step 4: Solve Part (c)

Express \(n (\sqrt[n]{a} - 1)\) as \(n (a^{1/n} - 1)\). Using the first-order approximation of \(a^{1/n} - 1 \approx \ln a / n\) for large \(n\), the product becomes \(\ln a\). Thus, \(\lim_{n \to \infty} n(a^{1/n} - 1) = \ln a\).

Step 5: Solve Part (d)

Rewrite \(n(\sqrt[n]{n} - 1)\) as \(n(e^{1/n \cdot \ln n} - 1)\). Approximate with \(e^{x} \approx 1 + x\) near zero, so it becomes \(n(\frac{\ln n}{n}) = \ln n\). As \(n \to \infty\), \(\ln n\) increases without bound. Thus, \(\lim_{n \to \infty} n(e^{1/n \cdot \ln n} - 1) = \infty\).

Key Concepts

Continuous Variables

Continuous variables are integral to understanding calculus and are often used in limit problems involving sequences. They differ from discrete variables as they can take on any value within a range, not just distinct, separate values.
Continuous variables are helpful in transforming problems, particularly in limits, where sequences naturally tend toward infinity.

• Continuous functions avoid abrupt changes, ensuring steady transitions between values.
• This makes limits with continuous variables predictable and follows smooth curves instead of jagged steps.

For example, when handling the exercise, replacing sequences involving integer variables like \( n \) with continuous counterparts such as \( x \) can make analysis more tractable.
This transformation often allows the application of continuous function properties and differentiation techniques.

L'Hôpital's Rule

L'Hôpital's Rule is an essential tool for evaluating limits that initially yield indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When these situations arise, derivatives provide a pathway to a solution.
In the textbook solution, part (b) demonstrated this rule:

• We started with \( \lim_{n \to \infty} \sqrt[n]{n} = \lim_{n \to \infty} e^{\frac{\ln n}{n}} \).
• The limit of the exponent, \( \frac{\ln n}{n} \), was of indeterminate form \( \frac{\infty}{\infty} \).
• Applying L'Hôpital's Rule required differentiating the numerator (\( \ln n \) leads to \( \frac{1}{n} \)) and the denominator (\( n \) becomes 1).
• This gave \( \lim_{n \to \infty} \frac{1/n}{1} = \lim_{n \to \infty} \frac{1}{n} = 0 \), simplifying the limit to give \( e^0 = 1 \).

Thus, L'Hôpital's Rule is an invaluable method of overcoming challenges with complex limits.

Logarithmic Functions

Logarithmic functions transform multiplicative relationships into additive ones. They are especially useful in solving limit problems involving exponential forms.
The logarithm introduces the possibility of simplifying problems using basic differentiation rules.

• This is apparent in parts (b) and (d) of the provided solution. Logarithms simplified the sequences \( \sqrt[n]{n} \) and \( \sqrt[n]{n} - 1 \) into more manageable forms, allowing differentiation.
• For example: \( \lim_{n \to \infty} \sqrt[n]{n} = \lim_{n \to \infty} e^{\frac{\ln n}{n}} \).

By rewriting expressions using logarithmic identities, they become approachable via derivative calculus and limit rules.
This illustrates why logarithms are indispensable when addressing complex limit expressions.

Infinite Limits

Infinite limits involve values growing without bound as they approach a particular threshold. Understanding infinite limits is crucial when evaluating expressions where sequences escalate infinitely.
In the exercise, infinite limits are primarily reflected in parts (c) and (d).

• For part (c), as \( n \rightarrow \infty \), roughly \( n(a^{1/n} - 1) \approx \ln a \).
• In part (d), the expression \( n(\sqrt[n]{n} - 1) \) equated to \( \ln n \), which grows indefinitely.

Recognizing infinite limits helps to identify asymptotic behavior or unbounded growth in functions.
Such insights are critical in calculus, enabling learners to navigate and comprehend sequences with infinite behavior effectively.