Problem 2 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

\(\lim _{x \rightarrow 0} \frac{x\left(e^{\left(\sqrt{1+x^{2}+x^{4}}-1\right) / x}-1\right)}{\sqrt{1+x^{2}+x^{4}}-1} \quad\) [Sep. 05, 2020 (II)] (a) is equal to \(\sqrt{e}\) (b) is equal to 1 (c) is equal to 0 (d) does not exist

Step-by-Step Solution

Answer

The limit is equal to 1, so the correct answer is (b).

Step 1: Express the Limit Problem

We have the limit problem to solve:\[ \lim _{x \rightarrow 0} \frac{x(e^{(\sqrt{1+x^{2}+x^{4}}-1)/x}-1)}{\sqrt{1+x^{2}+x^{4}}-1} \]

Step 2: Analyze the Expression under the Exponent

First, simplify the expression under the exponent:\[ \text{Expression: } e^{(\sqrt{1+x^{2}+x^{4}}-1)/x} \]For small values of \(x\), \(\sqrt{1+x^{2}+x^{4}} \approx 1 + \frac{x^2}{2} + \frac{x^4}{8} \) using a Taylor expansion for the square root.

Step 3: Simplify the Exponent

The expression \(\sqrt{1+x^{2}+x^{4}} - 1\) can be approximated for small \(x\):\[ \sqrt{1+x^{2}+x^{4}} - 1 \approx \frac{x^2}{2} + \frac{x^4}{8} \]Substituting, the exponent becomes:\[ \frac{(\frac{x^2}{2} + \frac{x^4}{8})}{x} = \frac{x}{2} + \frac{x^3}{8} \]

Step 4: Expand the Exponential Function

Now expand \(e^{\left(\frac{x}{2} + \frac{x^3}{8}\right)}\) using a Taylor series:\[ e^{\left(\frac{x}{2} + \frac{x^3}{8}\right)} = 1 + \frac{x}{2} + \frac{x^3}{8} + \frac{1}{2}\left(\frac{x}{2}\right)^2 + \dots \approx 1 + \frac{x}{2} + \text{higher order terms} \]

Step 5: Substitute Back and Simplify the Linear Fraction

Now substitute back into the original expression:\[ x \left( e^{\frac{x}{2}} - 1 \right) \approx x \left( 1 + \frac{x}{2} - 1 \right) = x \cdot \frac{x}{2} \approx \frac{x^2}{2}\]The denominator \(\sqrt{1+x^{2}+x^{4}}-1\) is already simplified as \(\frac{x^2}{2}\). Thus, the fraction is:\[ \frac{\frac{x^2}{2}}{\frac{x^2}{2}} = 1 \]

Step 6: Evaluate the Limit

As both the numerator and denominator simplify to equivalent expressions, the expression now is:\[ \lim_{x \to 0} \frac{x(e^{\frac{x}{2}} - 1)}{\sqrt{1+x^{2}+x^{4}}-1} = 1 \]

Key Concepts

Taylor Series Expansion

The Taylor series is a powerful method in calculus for approximating functions. Particularly useful for analyzing functions around a particular point, its general form involves an infinite sum of terms calculated from the values of a function's derivatives at a single point.
For the function in the original exercise, \(\sqrt{1 + x^2 + x^4}\), the Taylor series expansion helps approximate it when \(x\) is close to zero.
The expansion for the square root function relates as follows:

• First, realize the square root behaves like \(1 + \frac{x^2}{2} + \frac{x^4}{8}\) when expanded.
• Such expressions allow us to simplify and manipulate complex functions more easily.

This expansion allows you to break down complicated expressions into simpler components and solve problems that involve limits or continuity, like the one given in the problem.

Exponential Function

Exponential functions, often expressed as \( e^x \), are crucial in calculus and mathematical analysis due to their self-replicating derivative property.
To analyze their behavior for small \(x\), we utilize their Taylor series:
\( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \)
For a more focused approximation involving smaller terms, such as in the given expression \( e^{\left(\frac{x}{2} + \frac{x^3}{8}\right)} \), you may truncate the series:

• For small \(x\), higher order terms rapidly become negligibly small.
• In our exercise, we used these expansions to simplify \( e^{\frac{x}{2}} \) to \(1 + \frac{x}{2}\), ignoring higher degrees of \(x\).

This understanding allows you to transform complex limits into more manageable forms, crucial in limit evaluation.

Mathematical Analysis

Mathematical analysis involves delving into the core principles that govern calculus, including limits and continuity.
A major focus in analysis is understanding the behavior of functions as they approach certain points or infinity.
In our exercise, we use analysis by:

• Simplifying two complex expressions step-by-step using approximations.
• Utilizing the Taylor series to systematically handle terms in both the numerator and the denominator.

Such methods allow us to evaluate limits effectively. Here, post-simplification, the original limit problem finally simplifies entirely to 1 as \(x\) approaches zero.
The step-by-step process highlights the importance of reducing expressions into comparable parts, illustrating the essence of mathematical analysis.