Problem 80 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The limit is \(\ln x\), so the answer is (C).

Step 1: Simplify the Expression Inside the Limit

We start with the given expression inside the limit: \[x^{1/n} - x^{1/(n+1)}\]Apply the expression of power rules: \[x^{1/n} = e^{\ln(x)/n} \quad \text{and} \quad x^{1/(n+1)} = e^{\ln(x)/(n+1)}\]Use the first-order expansion of the exponential function, \(e^u \approx 1 + u\) for small \(u\). Thus, for large \(n\), \[x^{1/n} \approx 1 + \frac{\ln(x)}{n} \quad \text{and} \quad x^{1/(n+1)} \approx 1 + \frac{\ln(x)}{n+1}\]

Step 2: Simplify the Difference

Using the approximations from Step 1, the difference becomes: \[x^{1/n} - x^{1/(n+1)} \approx \left(1 + \frac{\ln(x)}{n}\right) - \left(1 + \frac{\ln(x)}{n+1}\right)\]This simplifies to: \[x^{1/n} - x^{1/(n+1)} \approx \frac{\ln(x)}{n} - \frac{\ln(x)}{n+1}\] Which further becomes: \[\frac{\ln(x)((n+1) - n)}{n(n+1)} = \frac{\ln(x)}{n(n+1)}\]

Step 3: Apply the Limit

Now, evaluate the limit:\[\lim_{n \to \infty} n^2 \left( x^{1/n} - x^{1/(n+1)} \right) \approx \lim_{n \to \infty} n^2 \cdot \frac{\ln(x)}{n(n+1)}\]Simplify the limit expression:\[= \lim_{n \to \infty} \frac{n^2 \ln(x)}{n(n+1)} = \lim_{n \to \infty} \frac{n \ln(x)}{n+1}\]For large \(n\), this expression simplifies to:\[= \lim_{n \to \infty} \ln(x) \cdot \frac{n}{n+1} = \ln(x) \cdot \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)\]Where the final limit is 1, therefore:\[= \ln(x)\]

Step 4: Conclusion

Based on our calculation, the limit evaluates to \( \ln(x) \). Thus, the correct answer to the given limit problem is option (C).

Key Concepts

Power Rules in Exponential Functions

When dealing with exponential functions, especially in limit problems, understanding how to manipulate powers can make complex expressions much simpler. The power rule in exponential functions helps you break down expressions like \(x^{1/n}\) into a form that can be more easily lhandled. For example, \(x^{1/n}\) can be rewritten as \(e^{\ln(x)/n}\), which shifts the focus from powers to exponents.

This transformation is useful because exponent expressions can be expanded using series approximations. As a power tends to infinity, the expression simplifies in a way that's more manageable using the properties of \(e\).

• Allows the transformation of the expression into a more straightforward form with \(e^{\ln(x)/n}\).
• Makes it easier to apply series expansions, particularly as \(n\) becomes large.

Essentially, the power rule clarifies complex transformations in limits, revealing underlying relationships between the variables involved.

First-order Expansion of Exponential Function

The exponential function \(e^u\) is approximately \(1 + u\) when \(u\) is small. This first-order expansion is incredibly helpful in calculus for simplifying expressions. And it's especially handy in limit problems where the variable approaches infinity or zero and makes the exponent small.

In our case, we used the approximation \(e^{\ln(x)/n} \approx 1 + \frac{\ln(x)}{n}\). As \(n\) becomes infinitely large, the exponent \(\ln(x)/n\) is quite small, making the first-order expansion valid.

• This method replaces complex exponential terms with simpler linear approximations.
• Allows for easy simplification and further calculation in limits and other problems.

Overall, first-order expansions serve as tools to unravel intricate exponential expressions, particularly in limit evaluations.

Properties of Natural Logarithms

Natural logarithms (\(\ln\)) have several properties that are useful for simplifying complex expressions. One key property is how they relate to exponential functions, which is crucial for dissecting terms like \(x^{1/n}\) into a manageable format. With the relation \(e^{\ln x} = x\), you can use logarithms to transform multiplication into addition and handle intricate calculations.

For solving limits involving natural logs, keep these properties in mind:

• The log of a power, \(\ln(x^a)\), becomes \(a \cdot \ln(x)\). This simplifies bases and makes differentiating or exploring behaviors easier.
• Logarithmic identities, like change of base formula, make it easy to work with different exponential forms.

In the context of limits like our exercise, the ability to decompose exponential expressions using natural logs lays the groundwork for approximations and ultimately, solving for the limit itself.