Problem 30 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(f(0)\) that makes the function continuous at \(x=0\) is \(\frac{1}{12}\). (Option A)

Step 1: Identify Function Behavior Near Zero

To determine if the function \(f(x)\) is continuous at \(x = 0\), we need to evaluate what happens to the function as \(x\) approaches zero. Specifically, we need to find \(\lim_{x \to 0} f(x)\).

Step 2: Express the Limit Formally

The limit of \(f(x) = \frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x}\) as \(x\) approaches zero can be written as:\[\lim_{x \to 0} \frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x}\]

Step 3: Use Series Expansion for the Numerator

Expand \(\sqrt[3]{1+x}\) and \(\sqrt[4]{1+x}\) using Taylor series near \(x = 0\): - \(\sqrt[3]{1+x} \approx 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \cdots\)- \(\sqrt[4]{1+x} \approx 1 + \frac{1}{4}x - \frac{1}{32}x^2 + \cdots\).

Step 4: Substitute the Expansions and Simplify

Substituting the expansions into the numerator:\[\sqrt[3]{1+x} - \sqrt[4]{1+x} \approx \left(1 + \frac{1}{3}x - \frac{1}{9}x^2\right) - \left(1 + \frac{1}{4}x - \frac{1}{32}x^2\right)\]This simplifies to:\[\frac{1}{3}x - \frac{1}{4}x + \left(\frac{1}{32} - \frac{1}{9}\right)x^2\]

Step 5: Simplify the Terms Further

Simplifying the terms from Step 4 gives:\[\frac{1}{3}x - \frac{1}{4}x = \frac{1}{12}x\]The \(x^2\) terms will become negligible as \(x\) approaches zero, focusing mainly on the linear term. Thus, as \(x \to 0\):\[\sqrt[3]{1+x} - \sqrt[4]{1+x} \approx \frac{1}{12}x\]

Step 6: Evaluate the Limit

Substitute back into the limit expression:\[\lim_{x \to 0} \frac{\frac{1}{12}x}{x} = \lim_{x \to 0} \frac{1}{12} = \frac{1}{12}\]

Step 7: Determine the Required Value at Zero

For the function \(f(x)\) to be continuous at \(x = 0\), \(f(0)\) must equal \(\lim_{x \to 0} f(x) = \frac{1}{12}\). Therefore, \(f(0) = \frac{1}{12}\).

Key Concepts

Limits and Continuity

When we talk about limits in mathematics, we're essentially trying to determine the value that a function approaches as the input (or "x" in most functions) gets closer to a certain point. Function continuity at a point is quite crucial. It's all about whether or not you can draw the graph of a function at that point without lifting your pencil. In simpler terms, a function is continuous at a point if there are no breaks, jumps, or holes in its graph at that point.
For a function to be continuous at a particular point, say at \( x = 0 \), the following condition needs to be satisfied:

• The function \( f(x) \) must be defined at \( x = 0 \)
• The limit of \( f(x) \) as \( x \) approaches 0 must exist
• The value of the function at \( x = 0 \), \( f(0) \), must equal the limit as \( x \) approaches 0

In our exercise, we verify this by determining that both the limit and \( f(0) \) should be equal to \( \frac{1}{12} \) for continuity.

Taylor Series Expansion

The Taylor Series is a mathematical tool used to approximate functions. It provides a way to express functions as the sum of an infinite series of terms that are calculated from the values of a function's derivatives at a single point. This is particularly useful for dealing with complicated functions that are difficult to comprehend at a glance.
In the case of the function given: \( f(x) = \frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x} \), we are interested in understanding its behavior around \( x = 0 \). This is where the Taylor Series shines.
For each part, \( \sqrt[3]{1+x} \) and \( \sqrt[4]{1+x} \), we expand them using Taylor series. These expansions help simplify the function and evaluate the limit by getting expressions like:

• \( \sqrt[3]{1+x} \approx 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \cdots \)
• \( \sqrt[4]{1+x} \approx 1 + \frac{1}{4}x - \frac{1}{32}x^2 + \cdots \)

By substituting these series expansions back into the function, we simplify the evaluation of the limit as \( x \) approaches zero and effectively handle both the function parts.

Mathematical Problem Solving

Solving mathematical problems, especially those involving limits and continuity, often requires a structured approach. Here's a simple breakdown of how to tackle such problems:First, identify the key point of interest. For the problem at hand, the focus is on making the function continuous at \( x = 0 \).
Next, use mathematical tools such as the Taylor series to simplify complex components of the function. Each term's simplification aids in understanding and evaluating the limit.
Then, apply limit laws and algebraic manipulation to find the limit of the function as \( x \) approaches the point of interest.
Finally, ensure the function's value at the point matches the computed limit. In our case, verifying that \( f(0) = \lim_{x \to 0} f(x) \) ensures continuity.
Mathematics is not just about numbers but about reasoning and process. Approaching problems systematically can significantly ease the complexity and enhance understanding.