Question

What is $\lim_{x \to 0} f(x)$, where $f(x) = x \sin \frac{1}{x}$?

• A. 0
• B. 1
• C. $\infty$
• D. Limit does not exist

Hint:

When a bounded function like $\sin \frac{1}{x}$ is multiplied by a term tending to $0$, the limit is $0$ by the squeeze theorem.

Answer 1

The Correct Option is A

Step 1: Recall the range of sine.
For all real numbers, $-1 \leq \sin \frac{1}{x} \leq 1$.
Step 2: Multiply by $x$.
Thus, $-x \leq x \sin \frac{1}{x} \leq x$.
Step 3: Apply the squeeze theorem.
As $x \to 0$, both $-x \to 0$ and $x \to 0$. Hence, by the squeeze theorem: \[ \lim_{x \to 0} x \sin \frac{1}{x} = 0. \]