Problem 45 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{p}=\lim _{\mathrm{x} \rightarrow 0^{+}}\left(1+\tan ^{2} \sqrt{\mathrm{x}}\right)^{\frac{1}{2 \mathrm{x}}}\) then \(\log \mathrm{p}\) is equal to : \([2016]\) (a) \(\frac{1}{2}\) (b) \(\frac{1}{4}\) (c) 2 (d) 1

Step-by-Step Solution

Answer

\( \log p = \frac{1}{2} \).

Step 1: Understand the Limit Expression

We need to evaluate \( \lim_{x \rightarrow 0^{+}}\left(1 + \tan^2 \sqrt{x}\right)^{\frac{1}{2x}} \). This expression contains a limit involving both an exponent and a trigonometric expression.

Step 2: Simplify the Trigonometric Expression

As \( x \to 0^+ \), \( \sqrt{x} \to 0^+ \) and \( \tan \sqrt{x} \approx \sqrt{x} \) because \( \tan t \approx t \) for small \( t \). Hence, \( \tan^2 \sqrt{x} \approx x \).

Step 3: Substitute the Approximation

Substitute \( \tan^2 \sqrt{x} \approx x \) into the original expression: \( 1 + \tan^2 \sqrt{x} \approx 1 + x \). Now the limit becomes \( \lim_{x \to 0^+} (1 + x)^{\frac{1}{2x}} \).

Step 4: Recognize the Exponential Limit Form

The expression \( (1 + u)^{\frac{1}{u}} \) as \( u \to 0 \) is a standard form that approaches \( e \). Here, set \( u = 2x \), so the expression becomes \( (1 + 2x)^{\frac{1}{2x}} \) and approaches \( e^{\frac{1}{2}} \) as \( 2x \to 0 \).

Step 5: Evaluate the Limit to Find \( p \)

Thus, the expression \( \lim_{x \to 0^+} (1 + x)^{\frac{1}{2x}} = e^{\frac{1}{2}} \), meaning that \( p = e^{\frac{1}{2}} \).

Step 6: Compute \( \log p \)

Compute \( \log p \): since \( p = e^{\frac{1}{2}} \), then \( \log p = \log(e^{\frac{1}{2}}) = \frac{1}{2} \log e = \frac{1}{2} \).

Key Concepts

Exponential Limits

Exponential limits are an important concept in calculus, particularly when dealing with expressions that involve an exponent in the context of a limit. When we have an equation of the form \((1 + u)^{1/u}\) as \(u\) approaches 0, it simplifies to the Euler's number \(e\), the base of natural logarithms. In many limit problems involving exponential expressions, we attempt to rewrite the expression in this standard form.

For example, let's analyze the original problem's limit expression: \(\lim_{x \to 0^+} (1 + x)^{\frac{1}{2x}}\). We can compare this to the form \((1 + u)^{1/u}\) by setting \(u = 2x\), which turns the expression into \((1 + 2x)^{\frac{1}{2x}}\). When \(x\) approaches zero, \(u\) also approaches zero, and the entire expression converges to \(e^{1/2}\).

Recognizing such patterns and knowing how exponential limits behave is crucial in calculus, particularly when analyzing behavior near points where functions are not well-defined.

Trigonometric Limits

Trigonometric limits often involve expressions with functions like sine, cosine, and tangent, and they appear frequently in limit problems. To solve trigonometric limits, it is important to remember some fundamental approximations. For small angles \(t\), \(\tan t \approx t\), and similarly, \(\sin t \approx t\). These approximations allow for simplification when taking limits.

In the given exercise, the expression \(1 + \tan^2 \sqrt{x}\) needed to be evaluated as \(x\) approaches zero. Because \(\tan t \approx t\) for small \(t\), \(\tan^2 \sqrt{x}\) approximates to \(x\) when \(x\) is near zero. The estimate \(1 + \tan^2 \sqrt{x} \approx 1 + x\) simplifies the problem into a form that can be handled using exponential limits, leading to the eventual solution.

Having a solid grasp of these basic trigonometric limit techniques is key to navigating more complex functions and understanding how they behave near critical points.

Differential Calculus

Differential calculus is a branch of calculus that focuses on rates of change and slopes of curves. While it is not directly used to solve the given limit problem, understanding differential calculus helps us appreciate the importance of limits. Limits are foundational in defining the derivative, which measures the rate at which a function is changing at any given point.

In essence, the derivative is the limit of the average rate of change of a function as the interval approaches zero. For instance, the derivative of a function \(f(x)\) is defined as \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).

Through understanding derivatives and their connection to limits, we are better equipped to grasp how functions behave, especially near points of interest like maxima, minima, or points of discontinuity. This knowledge is pivotal in calculus, providing tools to analyze and interpret real-world phenomena modeled by mathematical functions.