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

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

\(\log p = \frac{1}{2}\) which corresponds to (D).

Step 1: Analyze the Limit Expression

We need to find the value of \( p = \lim _{x \rightarrow 0+}\left(1+\tan ^{2} \sqrt{x}\right)^{\frac{1}{2x}} \). This expression suggests using the logarithm to simplify the limit. We set \( y = \left(1+\tan ^{2} \sqrt{x}\right)^{\frac{1}{2x}} \), and thus we need to evaluate \( \log p = \lim_{x \rightarrow 0+} \frac{1}{2x} \log \left(1 + \tan^2 \sqrt{x}\right) \).

Step 2: Apply Logarithmic Limit Transformation

Use properties of logarithms and limits to simplify: \( \log p = \lim_{x \rightarrow 0+} \frac{1}{2x} \log \left(1 + \tan^2 \sqrt{x}\right) \). For small values of \( u \), \( \log(1+u) \approx u \). Hence, \( \log \left(1 + \tan^2 \sqrt{x}\right) \approx \tan^2 \sqrt{x} \).

Step 3: Evaluate \(\tan^2 \sqrt{x}\) as \(x \rightarrow 0+\)

We have \( \tan^2 \sqrt{x} = \left( \frac{\sin \sqrt{x}}{\cos \sqrt{x}} \right)^2 = \frac{\sin^2 \sqrt{x}}{\cos^2 \sqrt{x}} \). For \(x\) approaching 0, \(\sin \sqrt{x} \approx \sqrt{x}\) and \(\cos \sqrt{x} \approx 1\), so \( \sin^2 \sqrt{x} \approx x \). Thus, \(\tan^2 \sqrt{x} \approx x\).

Step 4: Substitute Back to Limit Expression

Substitute \( \tan^2 \sqrt{x} \approx x \) back into the expression for \(\log p\), yielding \( \log p = \lim_{x \rightarrow 0+} \frac{x}{2x}\).

Step 5: Evaluate and Simplify the Limit

Simplify \( \log p = \frac{1}{2} \lim_{x \rightarrow 0+} \frac{x}{x} = \frac{1}{2} \times 1 = \frac{1}{2} \).

Key Concepts

Logarithmic Limit Transformation

When faced with complex limit expressions involving exponentials, a powerful method is the logarithmic limit transformation. This technique leverages the natural logarithm \(\log\) to simplify calculations and make problems more approachable. Let's explore how this works:

• In our original expression, \(p=\lim _{x \rightarrow 0+}\left(1+\tan ^{2} \sqrt{x}\right)^{\frac{1}{2x}}\), we transform it by taking the natural logarithm so that we can handle exponents more effectively.
• We define \(y = \left(1+\tan ^{2} \sqrt{x}\right)^{\frac{1}{2x}}\), meaning that \(\log p = \lim_{x \rightarrow 0+} \frac{1}{2x} \log \left(1 + \tan^2 \sqrt{x}\right)\). This allows us to work with the expression inside the limit rather than directly with the exponent.
• The approximation \(\log(1+u) \approx u\) for small \(u\) is key, as it transforms our expression into a simpler form that's easier to analyze.

This approach is particularly useful when the difficulty stems from expressions raised to powers as \(x\) approaches zero.

Trigonometric Limits

When evaluating limits involving trigonometric functions as the variable approaches a particular value, it's often helpful to use known approximations and transformations. A common technique is to replace trigonometric functions with simpler expressions.

• In this problem, we look at \(\tan^2 \sqrt{x} = \left( \frac{\sin \sqrt{x}}{\cos \sqrt{x}} \right)^2\). Near zero, these trigonometric functions can be approximated: \(\sin \sqrt{x} \approx \sqrt{x}\) and \(\cos \sqrt{x} \approx 1\).
• Using these approximations simplifies the computation, transforming \(\tan^2 \sqrt{x} \) to approximately \(x\).
• These approximations hold particularly well for small angles, capitalizing on the fact that small angle approximations are foundational in calculus and physics.

Understanding how to correctly apply and navigate these approximations can greatly streamline solving problems involving trigonometric limits.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for calculating limits that initially result in indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule provides a way to differentiate the numerator and denominator separately and then take the limit again.

• While L'Hôpital's Rule is not directly applied in our solution, understanding its scope can be valuable. In the case of our problem, the transformation of the limit into \ \frac{x}{x} \ simplifies the evaluation without needing further derivative work.
• However, should the need arise in more complicated expressions where a fraction form persists after simplification, applying L'Hôpital's Rule to each part could simplify the calculation.
• Overall, L'Hôpital's Rule assists in bypassing the initial indeterminate form by focusing on derivative behavior around the point of interest.

It’s essential to ensure that before using L'Hôpital's, the function is indeed indeterminate, and the conditions for the rule's application are properly met.