Problem 80 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty} n^{2}\left(x^{1 / n}-x^{1 / n+1}\right), x>0\) is equal to (A) 0 (B) \(e^{x}\) (C) \(\ln x\) (D) None of these

Step-by-Step Solution

Answer

(C) \( \ln x \)

Step 1: Rewrite the Expression

We start by examining the expression \( n^2(x^{1/n} - x^{1/(n+1)}) \). Notice that \( x^{1/n} \) and \( x^{1/(n+1)} \) are very close to each other when \( n \) is large. This suggests that the first-order approximation of these terms will be useful.

Step 2: Utilize Series Expansion

Using the property \( x^{1/n} = e^{(1/n) \ln x} \), we find the first-order Taylor expansion for small \( y \): \( e^y \approx 1 + y \). Apply it to both terms:- \( x^{1/n} \approx 1 + \frac{\ln x}{n} \)- \( x^{1/(n+1)} \approx 1 + \frac{\ln x}{n+1} \)

Step 3: Simplify the Difference

Now, calculate the expression \( x^{1/n} - x^{1/(n+1)} \):\[ (1 + \frac{\ln x}{n}) - (1 + \frac{\ln x}{n+1}) = \frac{\ln x}{n} - \frac{\ln x}{n+1} \]Simplify this to:\[ \frac{\ln x \cdot n+1 - n}{n(n+1)} = \frac{\ln x}{n(n+1)} \]

Step 4: Substitute Back into Limit

Substitute back into the limit:\[ \lim_{n \to \infty} n^2 \cdot \frac{\ln x}{n(n+1)} = \lim_{n \to \infty} \frac{n^2 \ln x}{n^2 + n} \]Simplify to:\\[ \ln x \cdot \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} \]

Step 5: Evaluate the Limit

As \( n \to \infty \), \( \frac{1}{n} \to 0 \). Thus, the limit becomes:\[ \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1 \]So, \( \ln x \cdot 1 = \ln x \).

Step 6: Conclusion

The original limit evaluates to \( \ln x \). Therefore, the correct answer is (C) \( \ln x \).

Key Concepts

Taylor Expansion

Taylor expansion is an extremely useful tool in calculus for approximating functions by polynomials near a certain point. The essence of a Taylor series is that it provides an infinite sum of terms derived from the function's derivatives at a single point. When dealing with small changes or perturbations, Taylor expansions give us approximate values in a manageable form.

In this particular exercise, we used a first-order Taylor expansion for the expression involving the terms \( x^{1/n} \). Since the exponent involves the variable \( n \) tending towards infinity, the change is small enough to apply a first-order expansion. This allowed us to approximate:

• \( x^{1/n} \approx 1 + \frac{\ln x}{n} \)
• \( x^{1/(n+1)} \approx 1 + \frac{\ln x}{n+1} \)

These approximations make it possible to simplify the calculation and reveal underlying patterns not easily spotted in the original complex form.

Series Expansion

Series expansions allow us to express complex functions as sums of simpler terms. This process is powerful for analyzing functions that seem hard to evaluate directly, especially when related to limits and large arguments.

In the problem at hand, the function \( x^{1/n} \) is expanded using the exponential behavior of powers. This expansion essentially views the function as a series to the leading terms, which are then easier to manage. The property \( e^y \approx 1+y \) for small \( y \) provides the groundwork:

• \( x^{1/n} = e^{(1/n) \ln x} \)
• \( e^y \approx 1 + y \) which leads to the approximations used

Through such expansions, problems about the behavior of functions as variables grow can be simplified significantly, allowing us to uncover solutions systematically.

Simplification Techniques

Simplification techniques are crucial when dealing with expressions that are cumbersome and difficult to manipulate. They make it easier to focus on the key aspects of a problem.

In the problem, the difference between \( x^{1/n} \) and \( x^{1/(n+1)} \) is calculated by realizing both terms are very close when \( n \) is large. Through simplification, the difference resolves to the expression \( \frac{\ln x}{n(n+1)} \), eliminating unnecessary complexity:

• We first expanded each term via the Taylor expansion.
• Then subtracted the two, noticing terms cancel out or simplify elegantly.

Such methods uncover the true nature of the expression, particularly valuable when approaching limits. As we substituted back into the limit, the overall expression became \( \ln x \), verifying the simplification was performed correctly.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as its arguments tend towards a specific limit, often infinity. Understanding this behavior provides insight, especially when approaching infinite limits, as in this exercise.

In this problem, recognizing the asymptotic relationship was key. As \( n \) becomes very large, terms like \( \frac{1}{n} \) go to zero. Evaluating such asymptotic expressions:

• Helps simplify complex limit problems by focusing on dominant terms.
• In this scenario, \( \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} \to 1 \).

Thus, observing how components vanish or stabilize informs us about the function's ultimate behavior. By analyzing asymptotic behavior, we accurately evaluated the limit and concluded that it equated to \( \ln x \).