Question

Let \( f, g : \mathbb{R} \to \mathbb{R} \) be two functions defined by \[ f(x) = \begin{cases} |x|^{1/8} \sin \tfrac{1}{|x|} \cos x & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \] and \[ g(x) = \begin{cases} e^x \cos \tfrac{1}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \] Then, which one of the following is TRUE?

• A. \( f \) is continuous at \( x = 0 \), and \( g \) is NOT continuous at \( x = 0 \)
• B. \( f \) is NOT continuous at \( x = 0 \), and \( g \) is continuous at \( x = 0 \)
• C. \( f \) is continuous at \( x = 0 \), and \( g \) is continuous at \( x = 0 \)
• D. \( f \) is NOT continuous at \( x = 0 \), and \( g \) is NOT continuous at \( x = 0 \)

Hint:

To check the continuity at a point, evaluate the limit of the function at that point. If the limit does not exist or does not match the function value, the function is not continuous.

Answer 1

The Correct Option is A

Step 1: Analyze the continuity of \( f(x) \).
To check if \( f(x) \) is continuous at \( x = 0 \), we need to evaluate the limit: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} |x|^{1/8} \sin \frac{1}{|x|} \cos x \] Since \( \sin \frac{1}{|x|} \) oscillates between -1 and 1 as \( x \to 0 \), and \( |x|^{1/8} \to 0 \), we have: \[ \lim_{x \to 0} f(x) = 0 \] However, \( f(0) = 0 \), so \( f(x) \) is continuous at \( x = 0 \). 
Step 2: Analyze the continuity of \( g(x) \).
We check if \( g(x) \) is continuous at \( x = 0 \): \[ \lim_{x \to 0} g(x) = \lim_{x \to 0} e^x \cos \frac{1}{x} \] Since \( e^x \) approaches 1 and \( \cos \frac{1}{x} \) oscillates, \( \lim_{x \to 0} g(x) \) does not exist, and \( g(x) \) is not continuous at \( x = 0 \). 
Final Answer: \[ \boxed{f \text{ is continuous at } x = 0, \, g \text{ is NOT continuous at } x = 0} \]