38. Find the limits of the following functions:
1) $f(x)=2 x-3-\frac{1}{x}$ as $x \rightarrow 1$:
2) $y=\frac{x^{3}-3 x^{2}+2 x-5}{x^{2}+2}$ as $x \rightarrow-1$;
3) $y=x \sin \frac{1}{x}$ as $x \rightarrow 0$.
Solution. Using the indicated theorems, we sequentially find:

1) $\lim _{x \rightarrow 1}\left(2 x-3-\frac{1}{x}\right)=\lim 2 \cdot \lim x-\lim 3-\frac{\lim 1}{\lim x}=$ $=2 \cdot 1-3-\frac{1}{1}=-2$
2) $\lim _{x \rightarrow-1} \frac{x^{3}-3 x^{2}+2 x-5}{x^{2}+2}=\frac{(\lim x)^{3}-3(\lim x)^{2}+2 \lim x-5}{(\lim x)^{2}+2}=$ $=\frac{(-1)^{3}-3(-1)^{2}+2(-1)-5}{(-1)^{2}+2}=\frac{-1-3-2-5}{3}=-\frac{11}{3}$;
3) as $x \rightarrow 0$, the argument $\frac{1}{x} \rightarrow \infty$, and the factor $\sin \frac{1}{x}$ will oscillate between -1 and +1, not tending to any specific number, i.e., this factor does not have a limit but is a bounded quantity, $\left|\sin \frac{1}{x}\right| \leqslant 1$. Therefore, according to Theorem II, the given function, representing the product of an infinitely small quantity $x$ and a bounded quantity $\sin \frac{1}{x}$, is an infinitely small quantity, and its limit is zero: $\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0$.