Find:

Solution (a) Since $\lim\limits_{x\rightarrow 0^{+}}x=0$ and $\lim\limits_{x\rightarrow 0^{+}}\ln x=-\infty$, then $x\ln x$ is an indeterminate form at $0^{+}$ of the type $0\cdot \infty$. We change $x\ln x$ to an indeterminate form of the type $\dfrac{\infty }{\infty }$ by writing $x\ln x= \dfrac{\ln x}{\dfrac{1}{x}}$ and using L’Hôpital’s Rule. $$\begin{eqnarray*} \lim\limits_{x\rightarrow 0^{+}}( x\ln x) &=&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\ln x}{\dfrac{1}{x}}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}\ln x}{\dfrac{d}{\textit{dx}}\dfrac{1}{x}} =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^{2}}}= \lim\limits_{x\rightarrow 0^{+}}( -x) =0\\[-20.9pt] &&\hspace{16pt}\quad\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}} {{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height13pt depth0pt}}\right.}}}\qquad\hspace{4.6pc} \hspace{3.7pc}\color{#0066A7}{\underset{\hbox{Simplify}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height13pt depth0pt}}\right.}}} \end{eqnarray*}$$

(b) Since $\lim\limits_{x\rightarrow \infty }\;x=\infty$ and $\lim\limits_{x\rightarrow \infty }\;\sin \dfrac{1}{x}=0$, then $x\sin \dfrac{1}{x}$ is an indeterminate form at $\infty$ of the type $0\cdot \infty$. We change $x\;\sin \dfrac{1}{x}$ to an indeterminate form of the type $\dfrac{0}{0}$ byeak writing $$\begin{eqnarray*} \lim\limits_{x\rightarrow \infty }x\;\sin \dfrac{1}{x}&=&\lim\limits_{x\rightarrow \infty } \dfrac{\sin \dfrac{1}{x}}{\dfrac{1}{x}}=\lim\limits_{t\rightarrow 0^{+}}\dfrac{\sin\;t}{t}=1 \\[-1.75pc] &&\qquad\hspace{26.1pt}\color{#0066A7}{\underset{{\hbox {Let} \;t=\dfrac{1}{x}}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}} \end{eqnarray*}$$