Problem 351 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{e^{\alpha x}-\cos \alpha x}{e^{\beta x}-\cos \beta x}\left\\{\text { Ans. } \frac{\alpha}{\beta}\right\\} $$

Step-by-Step Solution

Answer

To find the limit of the given function as x approaches 0, we apply L'Hôpital's Rule, which involves taking the derivatives of the numerator and denominator and then evaluating the limit of their quotient. We differentiate both the numerator and the denominator, obtaining the derivatives \(\alpha e^{\alpha x}+ \alpha \sin \alpha x\) and \(\beta e^{\beta x}+ \beta \sin \beta x\). Substituting these derivatives into the original limit, we get \(\lim_{x \rightarrow 0} \frac{\alpha e^{\alpha x}+ \alpha \sin \alpha x}{\beta e^{\beta x}+ \beta \sin \beta x}\). Evaluating the limit, we obtain \(\frac{\alpha}{\beta}\).

Step 1: Understanding L'Hôpital's Rule

L'Hôpital's Rule is a method used to evaluate the limits of functions that produce an indeterminate form such as 0/0 when directly applying the limit. In this case, as x approaches 0, both the numerator and denominator become indeterminate forms. To apply L'Hôpital's Rule, we take the derivative of the numerator and denominator with respect to the variable x, and find the limit of the quotient of these derivatives.

Step 2: Differentiate the numerator and denominator

Before applying L'Hôpital's Rule, we need to differentiate both the numerator and the denominator of the function with respect to x. Numerator: \(\frac{d}{dx}\left(e^{\alpha x}-\cos \alpha x\right) = \alpha e^{\alpha x}+ \alpha \sin \alpha x\) Denominator: \(\frac{d}{dx}\left(e^{\beta x}-\cos \beta x\right) = \beta e^{\beta x}+ \beta \sin \beta x\) Now we have the derivatives of the numerator and denominator with respect to x.

Step 3: Apply L'Hôpital's Rule

Now we apply L'Hôpital's Rule. We replace the original function with the quotient of their derivatives and find the limit as x approaches 0. \(\lim_{x \rightarrow 0} \frac{\alpha e^{\alpha x}+ \alpha \sin \alpha x}{\beta e^{\beta x}+ \beta \sin \beta x}\)

Step 4: Evaluate the limit

We'll now evaluate the limit as x approaches 0: \[\lim_{x \rightarrow 0} \frac{\alpha e^{\alpha x}+ \alpha \sin \alpha x}{\beta e^{\beta x}+ \beta \sin \beta x} = \frac{\alpha e^{0}+ \alpha \sin (0)}{\beta e^{0}+ \beta \sin (0)} = \frac{\alpha (1) + \alpha (0)}{\beta (1) + \beta (0)}\] Simplifying the expression, we get: \(\frac{\alpha}{\beta}\) So, the limit of the given function as x approaches 0 is \(\boxed{\frac{\alpha}{\beta}}\).

Key Concepts

Limit Evaluation

When we're exploring limit evaluation, especially with functions like \(\lim _{x \rightarrow 0} \frac{e^{\alpha x}-\cos \alpha x}{e^{\beta x}-\cos \beta x}\), our goal is to find the value that the function approaches as \(x\) gets very close to a certain point. Here, as \(x\) approaches 0, the function seems tricky because directly substituting \(x=0\) in the function gives us an indeterminate form like \(\frac{0}{0}\).
This is where techniques like L'Hôpital's Rule become invaluable. We have to manipulate the function, often through differentiation, to find a more straightforward expression that can be evaluated.

• Start by substituting the approaching value into the function.
• Identify if the limit leads to an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• If it does, use methods like L'Hôpital's Rule to simplify further.
• Finally, evaluate the simplified expression as \(x\) approaches the specified value.

Through these steps, we determine or 'evaluate' the limit of functions that exhibit complex behaviors near the point of interest.

Indeterminate Forms

Indeterminate forms can be quite perplexing in calculus. They occur when plugging the limiting values into the function results in expressions like \(\frac{0}{0}\), which do not clearly indicate the actual limit.
In our original problem, as \(x\) approaches 0, both the numerator and the denominator appear to result in 0, forming an indeterminate form.
Such scenarios require specialized techniques like L'Hôpital's Rule to handle.
Here’s a quick guide:

• Recognize the indeterminate forms (common ones include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\)).
• Apply L'Hôpital's Rule, which involves differentiating both the numerator and the denominator.
• Re-evaluate the limit using the derivatives.

Dealing with indeterminate forms requires patience and practice, as they form the stepping stones to mastering the art of limit evaluation.

Differentiation

Differentiation is a core tool in calculus, especially when dealing with limits involving indeterminate forms. In our problem, we must differentiate both the numerator and the denominator to apply L'Hôpital's Rule.
Let’s break it down:

• Numerator after differentiation: \(\frac{d}{dx}\left(e^{\alpha x}-\cos \alpha x\right) = \alpha e^{\alpha x}+ \alpha \sin \alpha x\)
• Denominator after differentiation:\(\frac{d}{dx}\left(e^{\beta x}-\cos \beta x\right) = \beta e^{\beta x}+ \beta \sin \beta x\)

Differentiation helps transform the original complicated limit into one that is simpler to evaluate.
Remember:

• Be familiar with derivative rules, especially for exponential functions and trigonometric functions.
• Carefully differentiate each term within the functions to ensure accurate calculations.
• Ensure all differentiation is performed with respect to the variable in question, typically \(x\).

By mastering differentiation, you'll have a solid foundation for tackling a wide range of calculus problems.

Exponential and Trigonometric Functions

Exponential and trigonometric functions frequently appear in calculus, playing significant roles in complex calculations. Understanding how they behave and how to differentiate them is vital.
For exponential functions like \(e^{\alpha x}\), the differentiation is straightforward:

• The derivative of \(e^{\alpha x}\) is \(\alpha e^{\alpha x}\), where \(\alpha\) is a constant that comes along due to the chain rule.

Trigonometric functions such as \(\cos \alpha x\) also follow specific rules:

• The derivative of \(-\cos \alpha x\) is \(\alpha \sin \alpha x\), again using the chain rule.

Knowing these rules allows us to differentiate terms quickly and confidently.
These functions combine within limits to form intriguing expressions that require careful analysis using tools like L'Hôpital's Rule.

• Familiarize yourself with the basic differentiation formulas of \(e^{x}\), \(\sin x\), and \(\cos x\).
• Understand how constants interact with these derivatives through the chain rule.

Effective application of these rules makes the process of solving calculus problems involving such functions both manageable and elegant.