Problem 271 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin x \cos x}\\{\text { Ans. } 2\\} $$

Step-by-Step Solution

Answer

The short answer is: \[ \lim_{x\rightarrow 0}\frac{e^x - e^{-x}}{\sin x\cos x} = 2 \]

Step 1: Understand the given limit expression

The given limit expression is: \[ \lim_{x\rightarrow 0}\frac{e^x-e^{-x}}{\sin x\cos x}. \]

Step 2: Apply the L'Hôpital's Rule

Since the limit is of the form \(\frac{0}{0}\), we can apply L'Hôpital's Rule, which states: \[ \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}, \] provided that \(\lim_{x\rightarrow a} f'(x)\) and \(\lim_{x\rightarrow a} g'(x)\) both exist and the limit on the right exists or is of the form \(\pm \infty\). Now let's find the derivatives of the numerator and denominator:

Step 3: Find the derivative of the numerator

To find the derivative of \(e^x - e^{-x}\), we can apply the chain rule. \[ \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x} \]

Step 4: Find the derivative of the denominator

To find the derivative of \(\sin x \cos x\), we can apply the product rule. \[ \frac{d}{dx}(\sin x\cos x) = (\cos x)(\cos x) - (\sin x)(\sin x)= \cos^2{x} - \sin^2{x}. \]

Step 5: Apply L'Hôpital's Rule

Replacing the given limit with the limit in L'Hôpital's Rule, we get: \[ \lim_{x\rightarrow 0}\frac{e^x - e^{-x}}{\sin x\cos x} = \lim_{x\rightarrow 0}\frac{e^x + e^{-x}}{\cos^2{x} - \sin^2{x}}. \]

Step 6: Evaluate the limit

As x approaches 0, we get the following: \[ \lim_{x\rightarrow 0}\frac{e^x + e^{-x}}{\cos^2{x} - \sin^2{x}} = \frac{e^0 + e^{-0}}{\cos^2{0} - \sin^2{0}} = \frac{2}{1-0} = 2. \] Thus, the value of the given limit is 2.

Key Concepts

Derivative

In calculus, understanding derivatives is key to working with functions and their rates of change. A derivative represents how a function value changes as its input changes. When dealing with the expression

• \(e^x - e^{-x}\)

, the derivative can be found using basic differentiation rules. For \(e^x\), the derivative is \(e^x\) itself, reflecting exponential growth. However, for \(e^{-x}\), the derivative turns into \(-e^{-x}\), because of the negative sign in the exponent, necessitating the chain rule.

By taking these derivatives, we find:

• \(\frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x}\)

, meaning the rate of change of the numerator reflects both exponential growth and decay in opposite directions. Grasping derivatives opens many doors in calculus. They provide insights into functions' behaviors and are foundational tools in evaluating limits and applying additional rules, such as L'Hôpital's Rule.

Limit Evaluation

Limit evaluation is a cornerstone of calculus, unraveling the behavior of functions as they approach specific points. In our exercise, we need to evaluate the limit:

• \(\lim_{x \rightarrow 0} \frac{e^x-e^{-x}}{\sin x \cos x}\)

. The direct substitution of \(x = 0\) yields the indeterminate form \(\frac{0}{0}\). This is where L'Hôpital's Rule becomes effective.

L'Hôpital's Rule allows us to differentiate the numerator and denominator separately and then re-evaluate the limit. With derivatives in hand, the exercise transitions from an indeterminate form to a more tangible expression. After computing new derivatives, we re-evaluate:

• \(\lim_{x\rightarrow 0}\frac{e^x + e^{-x}}{\cos^2{x} - \sin^2{x}}\)

, leading to \(\lim_{x\rightarrow 0}\frac{2}{1}\ = 2\). This reflects a common limit at the origin and showcases how powerful and enlightening the evaluation technique can be to simplify complex expressions.

Trigonometric Identities

Trigonometric identities are essential for simplifying and solving expressions involving trigonometric functions. In our limit problem, the denominator \(\sin x \cos x\) served a critical role before differentiation.

Through the use of product and sum-to-product trigonometric identities, we breakdown the expression during differentiation. Specifically, when differentiating \(\sin x \cos x\), utilizing the identity turns the product:

• \(\cos^2 x - \sin^2 x\)

, also known as the angle double formula: \(\cos(2x)\). Recognizing these identities allows us to transform complex expressions into simpler forms.

This understanding also aids in limit evaluation and contrasts trigonometric expressions with algebraic or exponential ones. Such techniques and identities underpin calculus and exemplify mathematical elegance in simplifying and solving real-world functional relationships.