# precise definition of a limit at infinity, application for limit at sin(x)

(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.

• For the definition, we want that given any $\epsilon\gt 0$ there is a $B$ such that if $x\gt B$ then $|f(x)-L|\lt \epsilon$. May 8, 2016 at 1:53
• Note that "so $|1/x-0|\lt 0$" does not make sense. But not great modification of your procedure will give a proof of the limit result. May 8, 2016 at 2:07
• Oh whoops I meant to put |1/x-0|<ϵ May 8, 2016 at 2:10
• Another minor thing, you write (almost) $1/(1/\epsilon)-0\lt \epsilon$, but it is actually equal to $\epsilon$. May 8, 2016 at 2:14
• For some basic information about writing math at this site see e.g. here, here, here and here. May 8, 2016 at 3:48

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 $t\ne 0$.