Problem 86 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The function \(F(x)\), defined as \(F(x)=\lim _{n \rightarrow \infty} \frac{f(x)+x^{2 n} g(x)}{1+x^{2 n}}\) shall be continuous everywhere, if (A) \(f(1)=g(1)\) (B) \(f(-1)=g(-1)\) (C) \(f(1)=-g(1)\) (D) \(f(-1)=-g(1)\)

Step-by-Step Solution

Answer

(A) and (B) ensure continuity.

Step 1: Understand the problem

The function \( F(x) \) is given by a limit expression. We need to identify the conditions under which this function remains continuous. Specifically, we need to identify the relationship between the functions \( f(x) \) and \( g(x) \) at the points \( x = 1 \) and \( x = -1 \) that ensures continuity.

Step 2: Analyzing the limit expression

The expression for \( F(x) \) is \[ F(x)=\lim _{n \rightarrow \infty} \frac{f(x)+x^{2 n} g(x)}{1+x^{2 n}}.\] As \( n \to \infty \), the term \( x^{2n} \) behaves differently based on \( x. \) At \( x = 1 \) or \( x = -1 \), \( x^{2n} = 1 \). This implies the behavior of \( f(x) \) and \( g(x) \) will affect the continuity at these points.

Step 3: Limiting behavior analysis at x = 1

Substituting \( x = 1 \) in the limit, we get \( F(1) = \lim _{n \rightarrow \infty} \frac{f(1) + g(1)}{2}. \) For \( F(x) \) to be continuous at \( x = 1 \), both expressions in the numerator must be equal: \( f(1) = g(1). \)

Step 4: Limiting behavior analysis at x = -1

Substituting \( x = -1 \) in the limit, we obtain \( F(-1) = \lim _{n \rightarrow \infty} \frac{f(-1) + g(-1)}{2}. \) For \( F(x) \) to be continuous at \( x = -1, \) it is required that \( f(-1) = g(-1). \)

Step 5: Evaluate the options

Given solutions are: (A) \( f(1)=g(1) \), (B) \( f(-1)=g(-1) \), (C) \( f(1)=-g(1) \), (D) \( f(-1)=-g(1) \). From our analysis, options (A) and (B) ensure the function remains continuous at the respective points \( x = 1 \) and \( x = -1 \). Therefore, the correct conditions are (A) \( f(1)=g(1) \) and (B) \( f(-1)=g(-1) \).

Key Concepts

Limits in Calculus

Limits form the groundwork of calculus and are essential in understanding how functions behave as they approach certain points. In the exercise, we are dealing with a limit involving a parameter growing towards infinity. The expression given is:

• \( F(x)=\lim_{n \to \infty} \frac{f(x)+x^{2n} g(x)}{1+x^{2n}} \)

This limit evaluates how the function \( F(x) \) behaves as the exponent \( n \) becomes infinitely large.

Understanding the behavior of this limit is crucial for solving the problem. As \( n \to \infty \), the term \( x^{2n} \) will determine which part of the numerator dominates. For either \( x = 1 \) or \( x = -1 \), the term \( x^{2n} \) simplifies to 1, simplifying our analysis of the function's limit at these points.

Thus, evaluating the limit is about understanding how the terms combine or simplify, helping us assess continuity for \( F(x) \) at these specific points.

Behavior of Functions at Specific Points

A function's behavior at specific points is vital in understanding its overall structure. In this problem's context, we focus on evaluating the function \( F(x) \) at points \( x = 1 \) and \( x = -1.\)

When evaluating the function at these critical points:

• At \( x = 1 \), substituting into the function gives us \( F(1) = \lim_{n \to \infty} \frac{f(1) + g(1)}{2} \). Here, both terms in the limit contribute equally, so continuity depends on these terms being equal, i.e., \( f(1) = g(1) \).


• Similarly, at \( x = -1, \) we derive \( F(-1) = \lim_{n \to \infty} \frac{f(-1) + g(-1)}{2} \). Again, the limit demands equality, \( f(-1) = g(-1) \) to ensure the function is continuous at this point.

Checking behavior at these key points aids in confirming the conditions for continuity by equating similar terms which simplify the evaluation.

Continuity Conditions

Continuity is key in ensuring a function behaves predictably without any jumps or gaps. For a function to be continuous at a point, its limit as it approaches that point must equal the function's value there.

In the given problem, continuity conditions are derived through the limit approach. At both \( x = 1 \) and \( x = -1 \), the conditions for continuity were:

• For \( x = 1: \) \( f(1) = g(1) \)
• For \( x = -1: \) \( f(-1) = g(-1) \)

If these conditions hold, the function \( F(x) \) defined by the limit behaves continuously at the specified points.

Assessing and applying these conditions ensures the transition of \( F(x) \) through \( x = 1 \) and \( x = -1 \) is seamless, guaranteeing no disruptions in the flow of the function's graph at these critical points. This understanding helps in constructing or analyzing continuous functions in calculus.