Problem 6 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(f(x)=\left\\{\begin{array}{cc}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.\), then \(\lim _{x \rightarrow 0} f(x)=\) (a) 1 (b) 0 (c) \(-1\) (d) None of these

Step-by-Step Solution

Answer

The limit \( \lim_{x \rightarrow 0} f(x) = 0 \), so the answer is (b) 0.

Step 1: Understanding the Problem

We need to find the limit of the function \( f(x) \) as \( x \) approaches 0. The function is defined piecewise as \( x \sin \frac{1}{x} \) when \( x eq 0 \) and 0 when \( x = 0 \).

Step 2: Applying Limit Laws

To solve \( \lim_{x \to 0} f(x) \), we'll first consider \( x \sin \frac{1}{x} \) as \( x \to 0 \). We need to use limit laws and properties of trigonometric functions to evaluate this.

Step 3: Boundaries of Sine Function

Since the sine function oscillates between -1 and 1, we can write: \( -1 \leq \sin \frac{1}{x} \leq 1 \). This property helps us to bound our original expression: \( -x \leq x \sin \frac{1}{x} \leq x \).

Step 4: Applying Squeeze Theorem

As \( x \to 0 \), both \( -x \) and \( x \) approach 0. By the Squeeze Theorem, since \( -x \leq x \sin \frac{1}{x} \leq x \), it follows that \( \lim_{x \to 0} x \sin \frac{1}{x} = 0 \).

Step 5: Finalizing the Limit

Since \( f(x) = x \sin \frac{1}{x} \) for \( x eq 0 \), and we found its limit to be 0, which matches \( f(0) = 0 \). Thus, \( \lim_{x \to 0} f(x) = 0 \). The answer is (b) 0.

Key Concepts

Piecewise functions

Piecewise functions are functions that are defined by different expressions depending on their input values. These functions can look a bit intimidating at first but they are not difficult to understand.
In the given exercise, the function \( f(x) \) is defined piecewise. This means:

• If \( x eq 0 \), then \( f(x) = x \sin \frac{1}{x} \).
• If \( x = 0 \), then \( f(x) = 0 \).

With piecewise functions, it's important to look at each piece separately. Understanding how the function behaves in different segments is crucial to determine the overall behavior of the function, particularly when evaluating limits. Such functions are commonly encountered in calculus as they reflect real-world scenarios where behaviors change suddenly at certain points.

Squeeze Theorem

The Squeeze Theorem is a powerful technique in calculus for finding limits of functions that might be otherwise hard to determine. It's also known as the Sandwich Theorem.The idea is simple. If you have a function that's hard to evaluate directly, but you know of two other functions that "squeeze" it from above and below, you can use these to determine its limit, provided they converge to the same limit.
For the function \( x \sin \frac{1}{x} \), it can be shown that:

• \( -x \leq x \sin \frac{1}{x} \leq x \).

Both the functions \( -x \) and \( x \) approach 0 as \( x \to 0 \). By the Squeeze Theorem, the function trapped between them, \( x \sin \frac{1}{x} \), must also approach 0 as \( x \to 0 \). This method is particularly helpful for functions involving trigonometry, where the behavior of oscillations can be neatly constrained.

Trigonometric limits

Trigonometric limits are frequently explored in calculus, given the oscillating nature of sine and cosine functions. These limits often require special attention to the properties of trigonometric functions.
One key property of the sine function is that it oscillates between -1 and 1, regardless of its argument. This makes trigonometric limits excellent candidates for approaches like the Squeeze Theorem. In particular, when dealing with expressions like \( \sin \left( \frac{1}{x} \right) \), knowing that \( -1 \leq \sin \frac{1}{x} \leq 1 \) can help in bounding the expression.When evaluating limits involving trigonometric functions, especially as they approach zero, it’s common to use approximations and limit laws related to trigonometric functions. With these, and techniques like the Squeeze Theorem, trigonometric limits often become manageable, allowing for successful evaluation in exercises like the given problem.