Question

For what values of \(\alpha\), the function \(f(x) = \sin x \cos\frac{1}{x}, x \ne 0\); \(f(x) = \alpha, x = 0\), is continuous at \(x = 0\)?

• A. 2
• B. 1
• C. 0
• D. 1/2

Hint:

The product of an infinitesimal function (a function whose limit is 0, like \(\sin x\)) and a bounded function (like \(\cos(1/x)\) or \(\sin(1/x)\)) always has a limit of 0 as the variable approaches the critical point.
This is a standard shortcut in calculus limits.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to find the value of the parameter \(\alpha\) that makes the piecewise function \(f(x)\) continuous at the point \(x = 0\).

Step 2: Key Formula or Approach:
For a function \(f(x)\) to be continuous at a point \(x = a\), the following mathematical condition must be satisfied:
\[ \lim_{x \to a} f(x) = f(a) \] In this problem, \(a = 0\). Therefore, we must have:
\[ \lim_{x \to 0} \left( \sin x \cos\frac{1}{x} \right) = f(0) = \alpha \]

Step 3: Detailed Explanation:
Let us evaluate the limit of the function as \(x \to 0\):
\[ L = \lim_{x \to 0} \sin x \cos\frac{1}{x} \] We can evaluate this limit using the Squeeze Theorem (also known as the Sandwich Theorem):
1. The cosine function is bounded between \(-1\) and \(1\) for all real arguments, regardless of how large or small they are. Therefore, for all \(x \ne 0\):
\[ -1 \le \cos\frac{1}{x} \le 1 \] 2. Since \(x \to 0\), we can assume \(x\) is in a small interval around \(0\) where \(\sin x \gt 0\) for \(x \gt 0\). Multiplying the entire inequality by the non-negative term \(\sin x\) (for \(x \gt 0\)) gives:
\[ -\sin x \le \sin x \cos\frac{1}{x} \le \sin x \] 3. Now, we take the limit of the outer bounding functions as \(x \to 0^+\):
\[ \lim_{x \to 0^+} (-\sin x) = 0 \] \[ \lim_{x \to 0^+} (\sin x) = 0 \] 4. Since both the lower limit and the upper limit converge to \(0\), the Squeeze Theorem guarantees that the limit of the middle function must also be \(0\):
\[ \lim_{x \to 0} \sin x \cos\frac{1}{x} = 0 \] 5. For the function to be continuous at \(x = 0\), this limit must equal \(f(0)\):
\[ L = \alpha \implies 0 = \alpha \] Thus, the function is continuous if and only if \(\alpha = 0\).

Step 4: Final Answer
The required value of \(\alpha\) is \(0\), which corresponds to option (C).