Question

If function \( f \) is defined as \[ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \] then prove that \( f \) is continuous.

Hint:

The Squeeze Theorem is an essential tool for finding limits of functions that involve products of a term that goes to zero (like \(x^2\)) and a bounded function (like \(\sin(\frac{1}{x})\) or \(\cos(\frac{1}{x})\)).

Answer 1

Step 1: Understanding the Concept:
A function \(f(x)\) is continuous at a point \(x=c\) if three conditions are met: \(f(c)\) is defined, \(\lim_{x \to c} f(x)\) exists, and \(\lim_{x \to c} f(x) = f(c)\). For the given piecewise function, we need to check continuity for \(x \neq 0\) and for \(x=0\).
Step 2: Key Formula or Approach:
For the case \(x=0\), we will use the Squeeze Theorem (or Sandwich Theorem) to evaluate the limit. The theorem states that if \(g(x) \leq f(x) \leq h(x)\) for all x in an open interval containing c (except possibly at c itself), and if \(\lim_{x \to c} g(x) = L = \lim_{x \to c} h(x)\), then \(\lim_{x \to c} f(x) = L\).
Step 3: Detailed Explanation:
Case 1: For \(x \neq 0\)
The function is defined as \(f(x) = x^2 \sin(\frac{1}{x})\). The functions \(x^2\), \(\frac{1}{x}\), and \(\sin(x)\) are all continuous on their respective domains. Since \(x \neq 0\), the composition \(\sin(\frac{1}{x})\) is continuous, and the product \(x^2 \sin(\frac{1}{x})\) is also continuous. So, \(f(x)\) is continuous for all \(x \neq 0\).
Case 2: For \(x = 0\)
We need to check if \(\lim_{x \to 0} f(x) = f(0)\). From the function definition, we know that \(f(0) = 0\).
Now, let's evaluate the limit: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \] We know that the sine function is bounded between -1 and 1, regardless of its input. \[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \quad (\text{for } x \neq 0) \] Multiply the entire inequality by \(x^2\). Since \(x^2 \geq 0\), the direction of the inequalities does not change. \[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \] Now, we take the limit of the bounding functions as \(x \to 0\): \[ \lim_{x \to 0} (-x^2) = 0 \] \[ \lim_{x \to 0} (x^2) = 0 \] Since both the lower and upper bounds approach 0, by the Squeeze Theorem, the function in the middle must also approach 0. \[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \] We have found that \(\lim_{x \to 0} f(x) = 0\) and we know \(f(0) = 0\). Since \(\lim_{x \to 0} f(x) = f(0)\), the function is continuous at \(x = 0\).
Step 4: Final Answer:
Since the function is continuous for all \(x \neq 0\) and also continuous at \(x=0\), we can conclude that the function \(f(x)\) is continuous for all real numbers. Hence proved.