Problem 58 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If function \(f(x)\) is differentiable at \(x=a\), then \(\lim _{x \rightarrow a} \frac{x^{2} f(a)-a^{2} f(x)}{x-a}\) is: \(\quad\) (a) \(-a^{2} f^{\prime}(a)\) (b) \(a f(a)-a^{2} f^{\prime}(a)\) (c) \(2 a f(a)-a^{2} f^{\prime}(a)\) (d) \(2 a f(a)+a^{2} f^{\prime}(a)\)

Step-by-Step Solution

Answer

The answer is (c) \(2af(a) - a^2 f'(a)\).

Step 1: Identify the Limit

We need to evaluate \( \lim _{x \rightarrow a} \frac{x^{2} f(a)-a^{2} f(x)}{x-a} \). This is a typical limit problem where substitution initially gives an indeterminate form \( \frac{0}{0} \).

Step 2: Rewrite the Expression

Notice that we can rewrite the limit as \( \lim _{x \rightarrow a} \frac{(x^2-a^2)f(a) - a^2(f(x) - f(a))}{x-a} \), which separates the parts involving \( f(a) \) and the difference \( f(x) - f(a) \).

Step 3: Factor and Simplify

Factor the term \( x^2 - a^2 \) as \((x-a)(x+a)\). The expression becomes \( \lim_{x \to a} \left( (x+a)f(a) - a^2 \frac{f(x)-f(a)}{x-a} \right) \).

Step 4: Apply Limit Properties

Observe that as \( x \to a \), the second term uses the definition of the derivative: \( \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a) \).

Step 5: Compute the Limit

Substitute back to evaluate the limit: 1. The first part \( \lim_{x \to a} (x+a)f(a) = 2af(a) \) as \( x \to a \). 2. The second part becomes \( -a^2 f'(a) \). Adding these results gives \( 2af(a) - a^2 f'(a) \).

Step 6: Match to the Options

Review the results. The expression \( 2af(a) - a^2 f'(a) \) matches option (c).

Key Concepts

Differentiability

Differentiability is an essential concept within calculus. In simple terms, a function is differentiable at a point if it has a defined derivative there. This indicates the function is smooth and has no sharp turns or cusps at that location.
A differentiable function will have the following attributes:

• Continuity: If a function is differentiable at a point, it is also continuous at that point.
• Smoothness: A differentiable function does not have abrupt changes in direction.

When solving problems that involve differentiability, it's crucial to remember that the derivative represents the rate at which the function is changing. To illustrate this, consider the function defined in the original exercise, where we need to find how the function changes as the input approaches a specific value, using the properties of differentiation.

Limit evaluation

Limits help us understand the behavior of functions as the input values approach a certain number. It's the backbone of derivatives and calculus as a whole. In many cases, evaluating a limit directly may lead to an indeterminate form like \( \frac{0}{0} \).
In the provided problem, limit evaluation involves the following:

• Identify the indeterminate form that arises from direct substitution.
• Apply algebraic manipulation such as factoring, splitting terms, or using conjugates, allowing us to simplify the expression further.

For example, in the original problem, factoring the expression \(x^2 - a^2\) was a key step in avoiding the indeterminate form and evaluating the limit successfully. Through this process, you can see how limits approach the values, indirectly leading to a neat solution.

Indeterminate forms

Indeterminate forms often come up in calculus, especially when dealing with limits. The most common indeterminate form is \( \frac{0}{0} \), but forms like \( \frac{\infty}{\infty} \), \(0 \cdot \infty\), and \(\infty - \infty\) also exist.
Dealing with these forms requires strategies such as L'Hôpital's Rule, algebraic manipulation, or series expansion. The key is to transform the expression into a form where standard limit rules or differentiation can be applied.
In our scenario, after encountering the \(\frac{0}{0}\) form, we modified the function, applied the definition of the derivative for the term \(f(x)-f(a)\), and used algebraic simplification. This helped transition from a problematic undefined form to a solvable limit. Understanding these techniques can enhance your problem-solving skills significantly when navigating calculus problems.