Problem 74 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{x \rightarrow 0} \frac{\left(2^{m}+x\right)^{1 / m}-\left(2^{n}+x\right)^{1 / n}}{x}\) is equal to (A) \(\frac{1}{m 2^{m}}-\frac{1}{n 2^{n}}\) (B) \(\frac{1}{m 2^{m}}+\frac{1}{n 2^{n}}\) (C) \(\frac{1}{m 2^{m-1}}-\frac{1}{n 2^{n-1}}\) (D) None of these

Step-by-Step Solution

Answer

(A) \( \frac{1}{m 2^m} - \frac{1}{n 2^n} \)

Step 1: Recognize the Function

The limit we need to evaluate is \( \lim_{x \rightarrow 0} \frac{(2^{m}+x)^{1/m}-(2^{n}+x)^{1/n}}{x} \). Notice that as \( x \rightarrow 0 \), the expressions under the fractional powers become close to \( 2^m \) and \( 2^n \) respectively. This suggests that we can use derivatives to simplify the expression.

Step 2: Apply the First Principles of Derivatives

The given limit has the form of the difference quotient that defines the derivative at a point. If we denote \( f(x) = (2^m + x)^{1/m} \) and \( g(x) = (2^n + x)^{1/n} \), then the derivative \( f'(x) \) at \( x=0 \) is \( f'(0) = \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x - 0} \).

Step 3: Differentiate the Functions

First, we differentiate \( f(x) = (2^m + x)^{1/m} \) using the chain rule: \[ f'(x) = \frac{1}{m} (2^m + x)^{\frac{1}{m} - 1} \cdot 1 = \frac{1}{m (2^m + x)^{1 - \frac{1}{m}}}. \]At \( x = 0 \), this becomes \( f'(0) = \frac{1}{m \cdot 2^m}. \)

Step 4: Differentiate the Second Function

Similarly, differentiate \( g(x) = (2^n + x)^{1/n} \):\[ g'(x) = \frac{1}{n} (2^n + x)^{\frac{1}{n} - 1} \cdot 1 = \frac{1}{n (2^n + x)^{1 - \frac{1}{n}}}. \]At \( x = 0 \), this becomes \( g'(0) = \frac{1}{n \cdot 2^n}. \)

Step 5: Substitute Derivatives into the Limit

The original limit becomes:\[ \lim_{x \rightarrow 0} \frac{f(x) - g(x)}{x} = f'(0) - g'(0). \]Substitute the derivatives we already calculated:\[ f'(0) - g'(0) = \frac{1}{m \cdot 2^m} - \frac{1}{n \cdot 2^n}. \]

Step 6: Choose the Correct Answer

The result matches option (A), which is \( \frac{1}{m \cdot 2^m} - \frac{1}{n \cdot 2^n} \).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. In simpler terms, the derivative is like a snapshot of a graph's slope at a particular moment.
For a function to have a derivative at a point means there's a defined tangent line at that point where we can determine the steepness.

• Geometrically, the derivative is like finding the slope of the tangent line to the curve of the function.
• Analytically, it's expressed as the limit of the average rate of change as the interval approaches zero.

Integrating derivatives allows us to solve many real-life problems, such as determining velocity, slope of a path, or tuning complex systems.
Understanding derivatives is crucial as they are applied in both simple and complex functions alike. They form the backbone of problems involving motion, where constant monitoring of change is necessary.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate compositions of functions. When you have a function inside another function, the chain rule lets you find the derivative of the entire composition.
Consider a composition of two functions:

• An outer function, say \( f \)
• An inner function, say \( g \)

The derivative of this composition is found by differentiating the outer function with respect to the inner function and then multiplying by the derivative of the inner function.
Mathematically, this is expressed as:\[(f(g(x)))' = f'(g(x)) imes g'(x)\]
The chain rule is often used in physics and engineering where systems are defined by nested functions. It enables practical evaluation of complex systems through simpler derivatives. Understanding the rule is essential for tackling more advanced calculus problems, as compositions appear frequently in various disciplines.

Difference Quotient

The concept of the difference quotient is central in understanding how derivatives are derived. It estimates the slope of the secant line through two points on a function's graph.
The difference quotient of a function \( f(x) \) at a point \( x \) is defined as:\[ \frac{f(x + h) - f(x)}{h}\]where \( h \) is a small increment approaching zero.

• It's essentially the average rate of change over a small interval.
• As \( h \) approaches zero, this quotient approaches the derivative of the function at that point.

In the solution of our original problem, the difference quotient represents the limit towards calculating the derivative, providing the connection from average to instantaneous rate of change.
Therefore, mastering the difference quotient is vital as it lays the groundwork for understanding derivatives and their practical applications in continuous change and real-world scenarios.