Problem 140 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(p=\lim _{x \rightarrow 0+}\left(1+\tan ^{2} \sqrt{x}\right)^{\frac{1}{2 x}}\) then \(\log p\) is equal to (A) \(\frac{1}{4}\) (B) 2 (C) 1 (D) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

\(\log p = \frac{1}{2}\), so the answer is (D).

Step 1: Understand the Expression Involved

We are given the limit \( p = \lim_{x \rightarrow 0+} \left(1+\tan^2 \sqrt{x}\right)^{\frac{1}{2x}} \). We need to find \( \log p \). Understanding this requires recognizing this expression as a limit that may lead to an indeterminate form, which suggests using the exponential and logarithms.

Step 2: Utilize Logarithmic Transformation

Take the logarithm of both sides to handle the complex power term. Set \( y = \log p = \lim_{x \rightarrow 0+} \frac{1}{2x} \log \left(1 + \tan^2 \sqrt{x} \right) \). This results from the property \( \log(a^b) = b \cdot \log a \).

Step 3: Simplify the Logarithm

For small values of \( \sqrt{x} \), \( \tan^2(\sqrt{x}) \approx (\sqrt{x})^2 = x \). So, \( 1 + \tan^2(\sqrt{x}) \approx 1 + x \). Therefore, \( \log(1 + \tan^2(\sqrt{x})) \approx \log(1 + x) \).

Step 4: Apply Expansion for Small \( x \)

Use Taylor expansion for \( \log(1 + x) \approx x \) for small \( x \). Then, \( \log(1 + \tan^2(\sqrt{x})) \approx x \). So, the expression becomes \( y = \lim_{x \rightarrow 0+} \frac{x}{2x} \).

Step 5: Evaluate the Limit

Simplify the expression \( \frac{x}{2x} = \frac{1}{2} \). Thus, \( \lim_{x \rightarrow 0+} \frac{x}{2x} = \frac{1}{2} \). Hence, \( \log p = \frac{1}{2} \).

Key Concepts

Indeterminate Forms

When faced with limits, sometimes the resulting expressions are "indeterminate forms." These forms, like \( \frac{0}{0} \) or \( \infty - \infty \), are expressions where the standard rules of limits don't easily apply. In this problem, as \( x \to 0^+ \), the expression involves \( \left(1 + \tan^2 \sqrt{x}\right)^{\frac{1}{2x}} \), leading to an indeterminate form of \( 1^{\infty} \). These forms are called indeterminate because they do not have a clear meaning without further manipulation.

Identifying an indeterminate form is the first step in determining how to manipulate the expression into a solvable format, typically using techniques like L'Hopital's Rule, Taylor Series, logarithmic transformations, or algebraic rearrangement. Recognizing these forms helps signal that a deeper examination is needed to fully evaluate the limit.

Logarithmic Transformation

Logarithmic transformation is a powerful technique for solving complex limit problems, especially those involving exponentiation. It simplifies handling expressions with powers by converting multiplicative relationships into additive ones.

In this exercise, taking the logarithm of both sides transforms the exponential form \((a^b)\) into \(b \cdot \log a\). For our limit, we rewrite \( p = \left(1 + \tan^2 \sqrt{x} \right)^{\frac{1}{2x}} \) as \( y = \lim_{x \to 0^+} \frac{1}{2x} \cdot \log \left(1 + \tan^2 \sqrt{x} \right) \). This transformation simplifies the evaluation by dealing with the exponential term in a much more manageable way.

Using logarithmic transformation, we transformed a problem initially involving an indeterminate exponential form into an expression that is easier to evaluate using known calculus techniques, such as Taylor expansion.

Taylor Expansion

The Taylor Expansion is a mathematical tool that approximates functions with polynomials. It is especially useful for dealing with functions when they go to zero or become "small" in value, and is derived from the function's derivatives at a particular point.

In our exercise, the expression \( \log(1 + \tan^2 \sqrt{x}) \) at \( x \to 0^+ \) can be approximated using the Taylor series for \( \log(1 + x) \), which is \( x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \), but for very small \( x \), it's enough to use the first term \( \log(1 + x) \approx x \). Hence, \( \log(1 + \tan^2 \sqrt{x}) \approx x \).

Applying this approximation, we simplify the limit problem further to \( y = \lim_{x \to 0^+} \frac{x}{2x} \). This expands into a straightforward calculation leading to the answer. The Taylor Expansion thus allows complex functions to be handled as simpler algebraic forms, crucial for evaluating limits of indeterminate forms.