Problem 84 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(\lim _{x \rightarrow \pi / 4} \frac{(\tan x)^{\tan x}-\tan x}{\ln (\tan x)-\tan x+1}\) is (A) \(-2\) (B) 1 (C) 0 (D) None of these

Step-by-Step Solution

Answer

The limit is \(-1\), which is not an option given (D).

Step 1: Identify Form of the Limit

We first recognize that the expression \(\lim _{x \rightarrow \pi / 4} \frac{(\tan x)^{\tan x} - \tan x}{\ln (\tan x) - \tan x + 1}\) involves both a subtraction of the same \(\tan x\) term and logarithmic function. As \(x\) approaches \(\pi/4\), \(\tan x\) approaches 1. Substituting 1 into both the numerator and the denominator gives \(0/0\), an indeterminate form, which suggests that we should use L'Hôpital's Rule after confirming the conditions are met.

Step 2: Apply L'Hôpital's Rule

Due to the \(0/0\) indeterminate form, we can use L'Hôpital's Rule, which involves differentiating the numerator and the denominator separately. First, differentiate the numerator: \[(\tan x)^{\tan x} = e^{\tan x \cdot \ln(\tan x)}.\] Differentiate using the chain and product rules to find: \[\frac{d}{dx}((\tan x)^{\tan x} - \tan x) = ((\tan x)^{\tan x})'(x) - \sec^2 x.\] For the denominator, simplify \[\ln(\tan x) - \tan x + 1\] differentiating gives \[\sec^2 x / \tan x - \sec^2 x.\]

Step 3: Differentiate the Numerator

Differentiate \((\tan x)^{\tan x} = e^{\tan x \ln(\tan x)}\). The derivative is: \[\frac{d}{dx}((\tan x)^{\tan x}) = ((\tan x)^{\tan x})(\ln(\tan x) + 1) \cdot \sec^2 x.\] Now combine the derivatives: \[((\tan x)^{\tan x})(\ln(\tan x) + 1) \cdot \sec^2 x - \sec^2 x.\]

Step 4: Differentiate the Denominator

Differentiating \(\ln(\tan x) - \tan x + 1\) gives: \[\frac{\sec^2 x}{\tan x} - \sec^2 x.\] At \( x \to \pi/4 \), this becomes a simpler expression that must be evaluated.

Step 5: Simplify and Evaluate the Limit

Using L'Hôpital's Rule, take the limit by substituting \(x \to \pi/4\) into the derivatives' ratio: \[\lim _{x \to \pi/4} \frac{((\tan x)^{\tan x})(\ln(\tan x) + 1) \cdot \sec^2 x - \sec^2 x}{\frac{\sec^2 x}{\tan x} - \sec^2 x}.\] Evaluate the constants and numeric values, where \(\tan(\pi/4) = 1\), making \(\ln(1) = 0\). This provides simplification to an expression without indeterminate form, yielding the evaluated limit as -(\tan(\pi/4)) = -1.

Key Concepts

Limits in Calculus

Limits in calculus are essential for understanding the behavior of functions as they approach a particular point. They tell us what value a function tends to as the input approaches a specific number. In the exercise provided, the limit is calculated as \( x \) approaches \( \pi/4 \). This means we are interested in what happens to the function \( \frac{(\tan x)^{\tan x} - \tan x}{\ln (\tan x) - \tan x + 1} \) when \( x \) gets very close to \( \pi/4 \).

When dealing with limits, it is common to encounter situations where simple substitution is not possible, especially if the result is undefined or indeterminate, like \( 0/0 \). This is where understanding the proper methods to evaluate these limits becomes crucial, including strategies like using L'Hôpital's Rule, which relies on derivative calculus to find meaningful limits.

Indeterminate Forms

Indeterminate forms arise in calculus when evaluating the limit of a function leads to ambiguous expressions such as \( 0/0 \), \( \infty/\infty \), or \( 0 \times \infty \), among others. These forms do not provide clear information about the limit's value, making special techniques necessary to resolve them.

In our given problem, substituting \( x = \pi/4 \) into the expression leads to a \( 0/0 \) indeterminate form. This signals that direct evaluation is not possible, and instead, techniques like L'Hôpital's Rule should be applied. L'Hôpital's Rule tells us to differentiate the numerator and the denominator separately and then take the limit again. This requires a good understanding of differentiation and sometimes multiple applications of the rule to resolve completely.

Understanding indeterminate forms is essential in calculus to interpret and solve problems involving limits correctly, especially when straightforward substitution does not provide usable results.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are fundamental in mathematics, especially in calculus as they often model periodic behavior. In the exercise at hand, the tangent function \( \tan x \) plays a significant role. As \( x \) approaches \( \pi/4 \), \( \tan x \) approaches 1 because \( \tan(\pi/4) = 1 \).

This understanding is key when trying to resolve the given limit problem, since knowing \( \tan(\pi/4) = 1 \) aids in simplifying the derivatives involved when applying L'Hôpital's Rule. Recognizing these landmark trigonometric function values can considerably simplify calculations and help navigate complex problems. Since the trigonometric functions also appear in the numerator and denominator, differentiating them properly is crucial in the limit evaluation.

Overall, a solid grasp of trigonometric functions and their properties is vital when tackling calculus problems involving limits, as is often the case with expressions that simplify using known trig identities.