(a) Write down the first principles definition of the statement $\lim\limits_{x→∞} f(x) = L$.
For this I have that for every $ε >0$, there is a corresponding number $N$, such that if $N>0$, then $|f(x)-L|<ε$.
(b) Using this definition, show that $\lim\limits_{x\to\infty} 1/x = 0$.
I have
$|1/x-0|<ε$
$1/|x|<ε$
$x>1/ε$
proof: If $x>1/ε$ then $|1/x-0|<|1/1/ε-0|<ε$ so $|1/x-0|<0$ and thus $\lim\limits_{x→∞} 1/x = 0$.
(c) Deduce from (b) and the fact that $\lim\limits_{t\to0} \sin(t) = 0$, that $\lim\limits_{x→∞} \sin(1/x) = 0$.
I'm a little stuck on this part. I know that $\sin(1/x)$ will give $0$ as $\lim_{t→0} \sin(t) = 0$ and $\lim_{x→∞} 1/x = 0$ so we'll just have $\sin(0) = 0$, but I'm unsure of how this can be worded.
Some items have been dealt with in comments, so we look only at c).
We want to show that for any $\epsilon\gt 0$, there is a $B$ such that if $x\gt B$ then $$|\sin(1/x)-0|\lt \epsilon.$$
Let $\epsilon\gt 0$. Since $\lim_{t\to 0}\sin t=0$ (given), there is a $\delta\gt 0$ such that if $0\lt |t-0|\lt \delta$, then $|\sin t-0|\lt \epsilon$.
Let $B=1/\delta$. If $x\gt B$, then $0\lt 1/x\lt \delta$, and therefore $|\sin(1/x)-0|\lt \epsilon$.
Remark: As the question asked, we assumed that $\sin t$ has limit $0$. We could dispense with that assumption by using the fact that $|\sin t|\lt |t|$ for all $t\ne 0$.