Question

Let \( f_i: \mathbb{R} \to \mathbb{R} \) for \( i = 1, 2 \) be defined as follows:
\[f_1(x) = \begin{cases} \sin\left(\frac{1}{x}\right) + \cos\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\0, & \text{if } x = 0 \end{cases}\]
and
\[f_2(x) = \begin{cases} x\left(\sin\left(\frac{1}{x}\right) + \cos\left(\frac{1}{x}\right)\right), & \text{if } x \neq 0 \\0, & \text{if } x = 0 \end{cases}\]
Then, determine the continuity of these functions at \( x = 0 \):

• A. \( f_1 \) is continuous at \( 0 \) but \( f_2 \) is not continuous at \( 0 \)
• B. \( f_1 \) is not continuous at \( 0 \) but \( f_2 \) is continuous at \( 0 \)
• C. Both \( f_1 \) and \( f_2 \) are continuous at \( 0 \)
• D. Neither \( f_1 \) nor \( f_2 \) is continuous at \( 0 \)

Answer 1

The Correct Option is B

For \( x \neq 0 \), \( f_1(x) = \sin\left(\frac{1}{x}\right) + \cos\left(\frac{1}{x}\right) \).

As \( x \to 0 \), \( \frac{1}{x} \to \infty \), and the terms \( \sin\left(\frac{1}{x}\right) \) and \( \cos\left(\frac{1}{x}\right) \) oscillate indefinitely without settling to any single value.

Thus, \( \lim_{x \to 0} f_1(x) \) does not exist.

Since \( \lim_{x \to 0} f_1(x) \neq f_1(0) \), \( f_1 \) is NOT continuous at \( x = 0 \).

Step 2: Analyze \( f_2(x) \):

For \( x \neq 0 \), \( f_2(x) = x \left(\sin\left(\frac{1}{x}\right) + \cos\left(\frac{1}{x}\right)\right) \).

As \( x \to 0 \), the oscillatory terms \( \sin\left(\frac{1}{x}\right) \) and \( \cos\left(\frac{1}{x}\right) \) remain bounded between \(-1\) and \(1\).

Therefore, \( |f_2(x)| \leq |x| \cdot (|\sin\left(\frac{1}{x}\right)| + |\cos\left(\frac{1}{x}\right)|) \leq 2|x| \), which tends to \( 0 \) as \( x \to 0 \).

Thus, \( \lim_{x \to 0} f_2(x) = 0 = f_2(0) \).

Hence, \( f_2 \) is continuous at \( x = 0 \).

Conclusion: \( f_1 \) is NOT continuous at \( 0 \), but \( f_2 \) is continuous at \( 0 \). Therefore, the correct answer is (B).