Problem 30 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(f(0)\) so that the function \(f(x)=\frac{\sqrt[3]{1+x}-\sqrt[4]{1+x}}{x}\) becomes continuous at \(x=0\), is (A) \(\frac{1}{12}\) (B) \(\frac{7}{12}\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The value of \( f(0) \) is \( \frac{1}{12} \) for continuity.

Step 1: Understand the problem

We need to find the value of \( f(0) \) such that the function \( f(x) = \frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x} \) is continuous at \( x = 0 \). For continuity at a point, the limit as \( x \to 0 \) must equal \( f(0) \).

Step 2: Setting up the Limit for Continuity

To find the limit of \( f(x) \) as \( x \rightarrow 0 \), consider the expression \( f(x) = \frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x} \). The limit will be evaluated by checking the behavior of the expression as \( x \to 0 \).

Step 3: Applying L'Hopital's Rule

Since \( \frac{0}{0} \) is an indeterminate form, we can apply L'Hopital's Rule. Differentiate the numerator and the denominator separately. For the numerator, find the derivatives of \( \sqrt[3]{1+x} \) and \( \sqrt[4]{1+x} \).

Step 4: Compute Derivatives

The derivative of \( \sqrt[3]{1+x} \) is \( \frac{1}{3}(1+x)^{-2/3} \), and the derivative of \( \sqrt[4]{1+x} \) is \( \frac{1}{4}(1+x)^{-3/4} \). So, the derivative of the numerator is \( \frac{1}{3}(1+x)^{-2/3} - \frac{1}{4}(1+x)^{-3/4} \).

Step 5: Evaluate the Limit

Substitute \( x = 0 \) in the derivative of the numerator \( \frac{1}{3}(1+0)^{-2/3} - \frac{1}{4}(1+0)^{-3/4} = \frac{1}{3} - \frac{1}{4} \). This simplifies to \( \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \).

Step 6: Conclusion on the Continuous Function

Thus, \( \lim_{{x \to 0}} f(x) = \frac{1}{12} \). Therefore, for \( f(x) \) to be continuous at \( x=0 \), we must have \( f(0) = \frac{1}{12} \).

Key Concepts

Limits and Continuity

In mathematics, understanding limits and continuity is crucial for analyzing the behavior of functions. A function is continuous at a point if there is no interruption or jump at that point. Its graph can be drawn without lifting the pen. Specifically, a function \( f(x) \) is said to be continuous at \( x = c \) if the following three conditions are met:

• \( f(c) \) is defined.
• The limit \( \lim_{{x \to c}} f(x) \) exists.
• The limit \( \lim_{{x \to c}} f(x) = f(c) \).

The problem at hand asks to determine the value of \( f(0) \) for which the given function is continuous at \( x = 0 \). This involves finding the limit of the function as \( x \to 0 \) and ensuring that this limit is equal to \( f(0) \). To handle situations where direct substitution results in undefined forms, such as \( \frac{0}{0} \), special techniques like L'Hopital's Rule and algebraic manipulation might be employed.

L'Hopital's Rule

L'Hopital's Rule is a valuable tool in calculus used to evaluate limits of indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When these forms appear, direct substitution into the limit expression does not provide an answer. Instead, we use L'Hopital's Rule which states:

• If \( \lim_{{x \to c}} f(x) = 0 \) and \( \lim_{{x \to c}} g(x) = 0 \), or both limits are infinite, then \( \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \), provided this latter limit exists.

In the exercise, the function is \( f(x) = \frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x} \), which forms an indeterminate \( \frac{0}{0} \) as \( x \to 0 \). By applying L'Hopital's Rule, we differentiate the numerator and the denominator individually to resolve this form and find the limit. This limit, \( \frac{1}{12} \), must equal \( f(0) \) for the function to be continuous at \( x = 0 \).

Indeterminate Forms

Indeterminate forms arise when trying to determine certain limits that do not have an immediate value from direct substitution. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \). These forms signal that we need to manipulate the expression differently to find a precise limit.In our problem, the expression \( \frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x} \) results in an indeterminate form \( \frac{0}{0} \) when \( x \) is substituted by 0. This occurs because both the numerator and the denominator approach zero as \( x \to 0 \). To address this, techniques such as factoring, rationalizing, or using calculus tools like L'Hopital's Rule are deployed. These methods transform the original expression into a form where a clear and definitive limit can be computed. In this case, L'Hopital's Rule was applied, yielding a continuous function at \( x=0 \) with \( f(0) = \frac{1}{12} \).