Answer
(a) $\lim\limits_{x \to \infty}f(x) = \lim\limits_{t \to 0^+}f(\frac{1}{t})$
and
$\lim\limits_{x \to -\infty}f(x) = \lim\limits_{t \to 0^-}f(\frac{1}{t})$
if these limits exist.
(b) $\lim\limits_{x \to 0^+}x~sin~\frac{1}{x} = 0$
Work Step by Step
(a) Suppose $\lim\limits_{x \to \infty}f(x) = c$
Let $\epsilon \gt 0$ be given.
There is a positive number $N$ such that $\vert f(x) - c\vert \lt \epsilon$ when $x \gt N$
Let $\delta = \frac{1}{N}$
Choose any $t$ such that $0 \lt t \lt \delta$
Then $\frac{1}{t} \gt \frac{1}{\delta} = N$
Then $\vert f(\frac{1}{t})-c \vert \lt \epsilon$
Therefore, $\lim\limits_{t \to 0^+}f(\frac{1}{t}) = c$
Now suppose $\lim\limits_{x \to -\infty}f(x) = c$
Let $\epsilon \gt 0$ be given.
There is a negative number $N$ such that $\vert f(x) - c\vert \lt \epsilon$ when $x \lt N$
Let $\delta = \frac{1}{N}$
Choose any $t$ such that $0 \gt t \gt \delta$
Then $\frac{1}{t} \lt \frac{1}{\delta} = N$
Then $\vert f(\frac{1}{t})-c \vert \lt \epsilon$
Therefore, $\lim\limits_{t \to 0^-}f(\frac{1}{t}) = c$
$\lim\limits_{x \to \infty}f(x) = \lim\limits_{t \to 0^+}f(\frac{1}{t})$
and
$\lim\limits_{x \to -\infty}f(x) = \lim\limits_{t \to 0^-}f(\frac{1}{t})$
if these limits exist.
(b) According to Exercise 65, $\lim\limits_{x \to \infty}\frac{sin~x}{x} = 0$
According to part (a):
$\lim\limits_{t \to 0^+}f(\frac{1}{t}) = \lim\limits_{x \to \infty} f(x)$
Therefore:
$\lim\limits_{t \to 0^+}t~sin~\frac{1}{t} = \lim\limits_{x \to \infty}\frac{sin~x}{x} = 0$
