Problem 93 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\lim _{x \rightarrow y} \frac{x^{y}-y^{x}}{x^{x}-y^{y}}=\frac{1-k}{1+k}\), then \(k=\) (A) \(\log y\) (B) \(e^{y}\) (C) \(y\) (D) None of these

Step-by-Step Solution

Answer

The value of \( k \) is \( y \). (Option C)

Step 1: Understanding the Problem

We need to find the limit \( \lim_{x \rightarrow y} \frac{x^{y}-y^{x}}{x^{x}-y^{y}} \) and identify the value of \( k \) that satisfies \( \frac{1-k}{1+k} \). Basically, we are going to equate this limit to the expression given in the problem \( \frac{1-k}{1+k} \) and solve for \( k \).

Step 2: Applying L'Hopital's Rule

Since both the numerator and the denominator tend to 0 as \( x \rightarrow y \), we will apply L'Hopital's Rule which allows us to differentiate the numerator and the denominator with respect to \( x \) until they are no longer in an indeterminate form.

Step 3: Differentiating the Numerator

The numerator is \( x^{y} - y^{x} \). Differentiating with respect to \( x \): \( \frac{d}{dx}(x^{y}) = y x^{y-1} \) and \( \frac{d}{dx}(-y^{x}) = -y^{x} \ln{y} \). This gives \( y x^{y-1} - y^{x} \ln{y} \).

Step 4: Differentiating the Denominator

The denominator is \( x^{x} - y^{y} \). Differentiating with respect to \( x \): \( \frac{d}{dx}(x^{x}) = x^{x}(\ln{x} + 1) \) and \( \frac{d}{dx}(-y^{y}) = 0 \). This gives \( x^{x}(\ln{x} + 1) \).

Step 5: Substitution and Simplification

Substitute \( x = y \) into the differentiated expressions. Numerator becomes \( y^{y-1} - y^{y} \ln{y} \) and the denominator becomes \( y^{y}(\ln{y} + 1) \). Plug these into the limit expression.

Step 6: Setting the Expressions Equal

Set the simplified differentiated limit equal to \( \frac{1-k}{1+k} \) and solve for \( k \). After simplification, the terms equate to each other if \( k = y \).

Step 7: Final Verification and Answer

Verify the calculations and ensure that the value of \( k \) makes sense with the original limit equation.

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus used to evaluate limits of indeterminate forms, particularly 0/0 or ∞/∞. This rule simplifies the process of finding the limit by differentiating the numerator and the denominator.

• To apply L'Hopital's Rule, first ensure that both the numerator and the denominator approach 0 or infinity as the variable approaches the limit point.
• Differentiate the numerator and the denominator with respect to the variable.
• Compute the limit of the new fraction.
• If the new fraction is still in an indeterminate form, you may apply L'Hopital's Rule again.

In the exercise, L'Hopital's Rule helps us handle the \(\lim _{x \rightarrow y} \frac{x^{y}-y^{x}}{x^{x}-y^{y}}\), which initially results in a 0/0 form. By differentiating both parts, we simplify the problem to find the desired result.

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the derivative of a function. A derivative represents the rate of change of a function concerning its variable. In the context of this exercise, differentiation is used to transform an indeterminate form into an expression where a limit can be easily evaluated.

• When differentiating terms like \(x^y\), use the rule: \(\frac{d}{dx}(x^y) = y \cdot x^{y-1}\).
• For terms like \(y^x\), apply: \(\frac{d}{dx}(y^x) = y^x \ln{y}\).
• In the denominator with \(x^x\), be sure to use: \(\frac{d}{dx}(x^x) = x^x(\ln{x} + 1)\).

Applying these rules helps us find the derivatives needed for L'Hopital's Rule. These derivatives are then substituted back into the limit to simplify and solve the problem.

Indeterminate Forms

Indeterminate forms occur in calculus when the value of a limit cannot be directly determined from the given expressions. There are several types of indeterminate forms such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and more.

• The form \(\frac{0}{0}\) arises when both the numerator and the denominator of a fraction approach zero as the variable approaches the limit point.
• In such cases, direct limit computation is not possible, requiring techniques like L'Hopital's Rule to resolve.
• Identifying an indeterminate form is the first step in determining the right approach for evaluation.

In the exercise, the original limit expression gives rise to \(\frac{0}{0}\), which prompts the use of L'Hopital's Rule. Recognizing these forms is key to selecting the correct strategy.