Problem 27 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

\(\lim _{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right)\) [Online May 7, 2012] (a) equals 1 (b) equals 0 (c) does not exist (d) equals \(-1\)

Step-by-Step Solution

Answer

The limit equals 0.

Step 1: Understanding the Problem

We are given the limit \( \lim_{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right) \) to solve. Our goal is to determine whether it equals 1, 0, does not exist, or is -1. The complication in this limit arises from the \( \sin\left(\frac{1}{x}\right) \) term, because \( \frac{1}{x} \) oscillates wildly as \( x \rightarrow 0 \).

Step 2: Analyze \( \frac{x - \sin x}{x} \)

The term \( \frac{x-\sin x}{x} \) can be approximated using Taylor expansion of \( \sin x \), which gives \( \sin x \approx x - \frac{x^3}{6} + \cdots \). Thus, \( x - \sin x \approx \frac{x^3}{6} \), leading to \( \frac{x-\sin x}{x} \approx \frac{x^2}{6} \). Therefore, as \( x \to 0 \), \( \frac{x - \sin x}{x} \to 0 \).

Step 3: Behavior of \( \sin\left(\frac{1}{x}\right) \)

The function \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 as \( x \rightarrow 0 \). Since it doesn't approach any fixed value, it makes the limit potentially undefined without understanding the factor it is multiplied by.

Step 4: Combine the Terms and Finally Evaluate the Limit

Combine the results from Steps 2 and 3. The term \( \frac{x - \sin x}{x} \) approaches 0, and when multiplied by \( \sin\left(\frac{1}{x}\right) \), which remains bounded between -1 and 1, the product will also be squeezed to 0. Hence, \( \lim_{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right) = 0 \).

Key Concepts

Limits

In calculus, limits help us understand the behavior of a function as the input approaches a particular value. It allows us to explore the value that the function is heading towards, even if the function does not explicitly reach or define that value.
A limit can be thought of as the intent or potential endpoint of a function's output. For example, even if a function doesn't successfully hit a target value due to an undefined point, limits let us analyze where the function is directed towards.
To calculate limits, we often rely on algebraic manipulation, substitution, or special techniques like L'Hôpital's Rule in cases of indeterminate forms.
In our current problem, we deal with a combination of terms that affect how we evaluate the limit as the input, in this case, moves toward zero. Understanding how these components influence each other is key to solving these types of problems.

Trigonometric Functions

Trigonometric functions, like sine, cosine, and tangent, are fundamental in calculus and represent periodic phenomena. The sine function, \(\sin x\), cycles between -1 and 1, which is crucial for problems where it appears in an infinite limit context.
A clear example is in the exercise given: \( \sin\left(\frac{1}{x}\right) \). As \( x \to 0 \), \( \frac{1}{x} \to \infty \), causing the sine function to oscillate between its limits indefinitely.
Understanding trigonometric function properties becomes essential in limits problems because their periodicity and bounded nature can greatly affect the outcome of the limit calculation, typically requiring additional tools like Taylor series or special limit rules to simplify the situations.

Taylor Series

A Taylor series is a powerful tool in calculus to approximate functions using an infinite sum of terms calculated from the values of a function’s derivatives at a single point. This approximation helps in simplifying complex functions to easier forms, which are then used for calculations like limits.
Consider the function \( \sin x \), which can be approximated as \( \sin x \approx x - \frac{x^3}{6} + \cdots \). Substituting this series expansion in place of \( \sin x \) allows us to linearize our function to better evaluate limits.
In the given exercise, using the Taylor series transforms \( x - \sin x \) into a less complex form, leaving behind higher-degree terms that tend towards zero as \( x \) approaches zero.
This form simplifies both the perception and calculation of limits, making Taylor series an indispensable part of analyzing and resolving otherwise difficult calculus problems.