limn →∞ n^2(x^1 / n-x^1 / n+1), x>0 is equal to (A) 0 (B) e^x (C) lnx (D) None of these

Name: Problem 80, Chapter 11: A Complete Resource Book in Mathematics for JEE Main
Uploaded: 2021-10-03T17:43:34-08:00
Duration: 2 min 56 s
Channel: Abhijith V
Description: limn →∞ n^2(x^1 / n-x^1 / n+1), x>0 is equal to (A) 0 (B) e^x (C) lnx (D) None of these

Step 1: We are given the limit $lim _{n rightarrow infty} n^{2}left(x^{1 / n}-x^{1 / n+1}right)$

Limn →∞ n^2(x^1 / n-x^1 / n+1), x>0 is equal to (A) 0 (B)...

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Key Concepts

Exponential Function and Its Series Expansion

The exponential function e^y can be approximated by its Taylor series when y is near zero. This is useful when dealing with expressions such as x^(1/n) which can be rewritten as e^(ln(x)/n) for large n. The Taylor series expansion allows for a simplification by considering only the dominant terms, making it easier to analyze limits and differences.

Taylor Series Expansion

The Taylor series provides a way to approximate functions near a specific point. When y is small, e^y can be approximated as 1 + y + y^2/2 + ... . In the context of the limit problem, by expanding x^(1/n) and x^(1/(n+1)) using the Taylor series, one can approximate the small difference between these terms and extract the leading order behavior.

Asymptotic Approximations and Leading Order Behavior

Asymptotic approximations involve finding the dominant terms in an expression as a parameter (like n) approaches infinity. By identifying and comparing the leading order terms in the Taylor expansions of similar expressions, one can determine the behavior of their difference multiplied by a function of n, as in the given limit. This technique is critical in evaluating limits where direct substitution is not feasible.

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