Problem 452 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

\text { If } \lim _{x \rightarrow 0} \frac{\\{(a-n) n x-\tan x\\} \sin x}{x^{2}}=0, \text { where } n \text { is non-zero real no., then find the value of } a

Step-by-Step Solution

Answer

The value of \(a\) is equal to \(n\).

Step 1: Use L'Hopital's Rule

This rule says we can take the derivative of the numerator and the denominator of a limit in the form \(0/0\) or \(∞/∞\) separately. The derivative of \(\{(a-n) n x-\tan x\} \sin x\) is \(\{(a - n)n - \sec^2x\}\sin x + \{(a - n) n x - \tan x\}\cos x\). The derivative of \(x^2\) is \(2x\).

Step 2: Reformulate the Limit

Apply the derivatives found from step 1 to the limit which now becomes \(\lim _{x \rightarrow 0} \frac{\{(a - n)n - \sec^2x\}\sin x + \{(a - n) n x - \tan x\}\cos x}{2x}\).

Step 3: Break Down to Recognizable Limits

Break down the above limit to \(\{(a - n)n \lim_{x \rightarrow 0} \frac{\sin x}{x} - \lim_{x \rightarrow 0} \frac{\sin x\sec^2x}{2x} + (a - n)n\lim_{x \rightarrow 0} \frac{x\cos x}{x} -\lim_{x \rightarrow 0} \frac{\tan x\cos x}{2}\} \).

Step 4: Solve the Limits

The first and third limits of step 3 are standard limits, thus evaluating to \(\{(a - n)n\times1 - 0 + (a - n)n - 0\}\) which gives \(2(a - n)n\).

Step 5: Setting the Limit to Zero

Since the problem states that the limit equals 0, then it must be true that \(2(a - n)n = 0\). Therefore, for valid non-zero real number \(n\), \(a = n\).

Key Concepts

L'Hopital's Rule

In calculus, when dealing with limits, you might encounter indeterminate forms such as

• \(0/0\)
• \(∞/∞\)

These forms are tricky. That's where L'Hopital's Rule comes to the rescue. This rule allows us to find limits of these indeterminate forms by differentiating the numerator and the denominator separately. Keep in mind that this rule can be applied repeatedly until the indeterminate form is resolved. In our original problem, the limit was in the form of \(0/0\), which made it perfect for using L'Hopital's Rule. By differentiating both parts, we simplify the problem into a more manageable form.
L'Hopital's Rule can make apparently difficult problems much easier by transforming a limit problem based on differentiation.

Limits

The concept of limits is fundamental in calculus. A limit represents what a function approaches as the input approaches some value. Limits help
to define derivatives and are crucial when dealing with continuous functions. A typical
limit problem involves finding the behavior of a function as the input gets very close toIn the initial problem, the limit expression was complex, but breaking it down into simpler components made it easier to handle.
Recognizing standard limits, such as \(\lim_{x \to 0}\frac{\sin(x)}{x} = 1\), was crucial in finding the final result.
Limits provide the foundation for future calculus concepts like derivatives and integrals.

Trigonometric Limits

Trigonometric limits involve understanding and evaluating limits that include trigonometric functions like sine, cosine, and tangent. Their behavior, particularly as they approach zero or infinity, often involves certain patterns that are widely recognized in calculus:

• For instance, \(\lim_{x \to 0}\frac{\sin(x)}{x} = 1\) is a classic example of a trigonometric limit.
• Another is \(\lim_{x \to 0}\frac{1 - \cos(x)}{x^2} = \frac{1}{2}\).

Understanding these patterns can help simplify problematic expressions.Trigonometric limits often arise in physics and engineering, where periodic functions describe wave patterns. In the problem, recognizing and using these standard limits helped simplify a complex expression and reach the solution efficiently.

Derivatives

Derivatives are the building blocks of calculus. They represent the rate of change of a function concerning its variable. By finding the derivative, you can determine how a slight change in the variable affects the function's outcome. This concept was critical in the original problem when differentiating both the numerator and denominator of our limit expression.
The derivative rules, such as the product rule and chain rule, were instrumental when we differentiated expressions involving products of functions and compositions of functions. Understanding derivatives allows us to tackle complex calculus problems by breaking them down into smaller, more manageable parts.
In real life, derivatives have applications in areas such as physics, where they represent velocity and acceleration.