To find the value of the limit [tex]\(\lim _{x \rightarrow 0} x \sin \frac{1}{x}\)[/tex], let's analyze the behavior of the function as [tex]\(x\)[/tex] approaches 0.
1. Expression: The function given is [tex]\(x \sin \frac{1}{x}\)[/tex].
2. Bound the Sine Function: We know that the sine function oscillates between -1 and 1. Therefore, for any value of [tex]\(y\)[/tex],
[tex]\[
-1 \leq \sin(y) \leq 1.
\][/tex]
Specifically, for [tex]\(y = \frac{1}{x}\)[/tex],
[tex]\[
-1 \leq \sin \frac{1}{x} \leq 1.
\][/tex]
3. Multiply by [tex]\(x\)[/tex]: When we multiply all parts of the inequality by [tex]\(x\)[/tex], we obtain
[tex]\[
-x \leq x \sin \frac{1}{x} \leq x.
\][/tex]
4. Behavior as [tex]\(x\)[/tex] Approaches 0:
- As [tex]\(x\)[/tex] approaches 0 from the positive side ([tex]\(x \rightarrow 0^+\)[/tex]), [tex]\(x\)[/tex] becomes very small and positive.
- As [tex]\(x\)[/tex] approaches 0 from the negative side ([tex]\(x \rightarrow 0^-\)[/tex]), [tex]\(x\)[/tex] becomes very small and negative.
5. Limit Analysis: Consider the inequality:
[tex]\[
-|x| \leq x \sin \frac{1}{x} \leq |x|.
\][/tex]
Here, [tex]\(|x|\)[/tex] denotes the absolute value of [tex]\(x\)[/tex], which is just [tex]\(x\)[/tex] if [tex]\(x\)[/tex] is positive and [tex]\(-x\)[/tex] if [tex]\(x\)[/tex] is negative. In either case, as [tex]\(x\)[/tex] approaches 0, [tex]\(|x|\)[/tex] also approaches 0.
6. Squeeze Theorem: According to the Squeeze Theorem, if
[tex]\[
-|x| \leq x \sin \frac{1}{x} \leq |x}
\][/tex]
and both [tex]\(-|x|\)[/tex] and [tex]\(|x|\)[/tex] converge to 0 as [tex]\(x \rightarrow 0\)[/tex], then
[tex]\[
\lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0.
\][/tex]
Thus, the limit of the function [tex]\(x \sin \frac{1}{x}\)[/tex] as [tex]\(x\)[/tex] approaches 0 is
[tex]\[ \boxed{0}. \][/tex]
