Use the Squeeze Theorem to find $\lim\limits_{x\rightarrow 0}\left( x\sin \dfrac{1}{x}\right)$.

Solution If $x\neq 0$, then $\sin \dfrac{1}{x}$ is defined. We seek two functions that “squeeze” $y=x\sin \dfrac{1}{x}$ near $0.$ Since $-1\leq \sin x\leq 1$ for all $x,$ we begin with the inequality $$\begin{equation*} \left\vert \sin \dfrac{1}{x}\right\vert \leq 1\qquad x\neq 0 \end{equation*}$$

Since $x\neq 0$ and we seek to squeeze $x\sin \dfrac{1}{x},$ we multiply both sides of the inequality by $\vert x\vert ,$ $x\neq 0.$ Since $\vert x\vert >0,$ the direction of the inequality is preserved. Note that if we multiply $\left\vert \sin \dfrac{1}{x}\right\vert \leq 1$ by $x,$ we would not know whether the inequality symbol would remain the same or be reversed since we do not know whether $x>0$ or $x\lt 0.$ $$\begin{equation*} \begin{array}{r@{\qquad}ll} |x|\left\vert \sin \dfrac{1}{x}\right\vert \leq |x| & & {\color{#0066A7}{\hbox{Multiply both sides by \(\vert x \vert >0\).}}} \\[12pt] \left\vert x\sin \dfrac{1}{x}\right\vert \leq |x| & & {\color{#0066A7}{{|a|\cdot |b| = |a\,b|}.}} \\[11pt] -\vert x\vert \leq x\sin \dfrac{1}{x}\leq |x| & & {\color{#0066A7}{\vert a\vert\,{\leq}\,b\hbox{ is equivalent to } {-b\leq a\leq b.}}} \end{array} \end{equation*}$$

Figure 39 $g(x)=x\sin\dfrac{1}{x}$ is squeezed between $f(x)=-\vert x\vert$ and $h(x)=\vert x\vert$.

Now use the Squeeze Theorem with $f(x) = - |x|$, $g(x) = x\sin \dfrac{1}{x}$, and $h(x) = |x|$. Since $f(x)\leq g(x)\leq h(x)$ and $$\begin{equation*} \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}(-|x|)=0\qquad \hbox{and}\qquad \lim_{x\rightarrow 0}h(x)=\lim_{x\rightarrow 0}\vert x\vert =0 \end{equation*}$$

it follows that $$\lim_{x\rightarrow 0}g(x)=\lim_{x\rightarrow 0}\left( x\cdot \sin \frac{1}{x} \right) =0$$