Problem 121 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The function \(f: R \sim\\{0\\} \rightarrow R\) given by \(f(x)=\frac{1}{x}-\frac{2}{e^{2 x}-1}\) can be made continuous at \(x=0\) by defining \(f(0)\) as [2007] (A) 2 (B) \(-1\) (C) 0 (D) 1

Step-by-Step Solution

Answer

The correct value to define \(f(0)\) for continuity is 0. (Option C)

Step 1: Clarify the Problem

We are given a function \(f(x)\) that is not defined at \(x=0\). The goal is to find a value for \(f(0)\) that makes the function continuous at that point.

Step 2: Apply Continuity Definition

For a function \(f(x)\) to be continuous at \(x = 0\), the limit of \(f(x)\) as \(x\) approaches 0 must be equal to \(f(0)\). Thus, we need to calculate \(\lim_{x \to 0} f(x)\).

Step 3: Calculate the First Part of the Function

Consider the term \(\frac{1}{x}\). As \(x\to 0\), \(\frac{1}{x}\) tends to \(\pm \infty\), depending on the direction from which \(x\) approaches zero. However, our interest is in the behavior close to zero from both directions.

Step 4: Evaluate the Second Part of the Function

For the term \(-\frac{2}{e^{2x}-1}\), using a first-order expansion of \(e^{2x}\) around \(x = 0\), we have \(e^{2x} \approx 1 + 2x\). Thus, \(e^{2x} - 1 \approx 2x\). It follows that \(-\frac{2}{e^{2x}-1} \approx -\frac{2}{2x} = -\frac{1}{x}\).

Step 5: Combine Terms and Simplify

Combine the two approximated terms: \(\lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x} \right) = \lim_{x \to 0} 0 = 0\). Therefore, \(\lim_{x \to 0} f(x) = 0\).

Step 6: Select the Correct Value for Continuity

Since \(\lim_{x \to 0} f(x) = 0\), we can define \(f(0) = 0\) to make the function continuous at \(x = 0\). The answer is option (C), 0.

Key Concepts

Limit Calculation

Calculating the limit of a function as it approaches a specific value is crucial in understanding the function's behavior at that point. For the function \[ f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1} \] we need to calculate the limit as \( x \to 0 \). To say a function is continuous at \( x = 0 \), the limit \[ \lim_{x \to 0} f(x) \] must exist and be equal to \( f(0) \). The process involves simplifying the terms, and since \( \frac{1}{x} \) explodes to infinity as \( x \to 0 \), we must ensure careful handling by combining it with compensating terms. Ensuring the components of the function cancel correctly can reveal the true limit.

• Identify potential discontinuities.
• Simplify using limit rules and algebraic manipulation.
• Check if limit equals the defined value for continuity.

Indeterminate Forms

Indeterminate forms arise in calculus when evaluating limits; expressions seem irreducible at first glance. One common indeterminate form is \[ \frac{0}{0} \]which often appears during limit evaluations of functions. In our function, as \( x \to 0 \), both terms, \( \frac{1}{x} \) and \( -\frac{2}{e^{2x}-1} \), lead to this format when combined. To resolve it, we employ algebraic manipulations and asymptotic expansions:

• For \( \frac{1}{x} \), individually diverges.
• For \(-\frac{2}{e^{2x}-1} \), approaches \( -\frac{1}{x} \).
• Together, they simplify to \( 0 \) facilitating continuity.

Indeterminate forms often require methods like L'Hôpital's Rule or algebraic simplification to resolve, ensuring we understand the function's deeper behavior around challenging points.

Exponential Expansion

Exponential expansion is a technique to approximate functions involving exponential terms, often using Taylor or Maclaurin series. For small values of \( x \), the exponential function \[ e^{2x} \]can be expanded via its series as \[ e^{2x} \approx 1 + 2x + (2x)^2/2! + \dots \]For our problem, we simplified using just the first-order approximation,\[ e^{2x} \approx 1 + 2x \]. By substituting this into the denominator, \( e^{2x} - 1 \approx 2x \), allowing us to conclude\(-\frac{2}{e^{2x}-1} \approx -\frac{1}{x} \). This approximation is pivotal, enabling us to resolve the indeterminate forms around \( x = 0 \) and ensuring the function behaves predictably:

• Simplifies complex exponential elements.
• Makes solving limits feasible.
• Supports continuity evaluations.

When dealing with exponential functions near zero, utilizing appropriate expansions can make seemingly complex equations manageable.