QUESTION: Let $f$ be a continuous function from $\Bbb{R}$ to $\Bbb{R}$ (where $\Bbb{R}$ is the set of all real numbers) that satisfies the following property:
For every natural number $n$, $f(n) =$ the smallest prime factor of $n.$ For example, $f(12) = 2, f(105) = 3.$ Calculate the following-
$(a)\lim_{x→∞}f(x)$
$(b)$ The number of solutions to the equation $f(x) = 2016$.
MY SOLUTION: I do not have a problem in understanding part $(a)$. Clearly, $\infty$ is not a number and we cannot find any smallest prime factor for it. Or we can also argue that for any even number-
$f(even)=2$
And for any odd number it depends on the type of the odd number.. in case it's prime then $f(prime \space x)=x$ and in case it's not prime then the answer is something else..
Anyway, we find that there are numerous possibilities and since all of these possibilities directly depend on the number we have chosen so, we cannot account for what happens in the case of $\infty$.
Coming to the second part, the question itself bounced over my head. Let's see carefully what it says-
We know, $f(x)=$ the smallest prime factor of $x$. Therefore, $f(x)=2016$ must imply (by the same logic that)-
The smallest prime factor of $x$ is $2016$.
Wait. What?! Firstly, 2016 is not prime. So how can I account for $x's$ which have such an impossible prime factor.. secondly, even if we assume that $2016$ is the smallest factor of $x$ there are infinite $x's$ which satisfies such a property. Our answer in that case is not bounded (although it's nowhere mentioned it should be)..
So what does the second part even mean? 🤔
Thank you in advance for your kind help :).
Best Answer
For a, you can conclude that the limit does not exist. As you say, $f(n)=2$ for even $n$ and $f(n) \ge 3$ for odd $n$. If you think of the $N-\epsilon$ definition of a limit at infinity, this function will fail to have a limit and you can choose any $\epsilon \lt \frac 12$ to demonstrate that.
For b, you are expected to use the fact that $f(x)$ is continuous and use the intermediate value property. We have $f(2016)=f(2018)=2$ and $f(2017)=2017$ because $2017$ is prime. There must be at least one number in the intervals $(2016,2017)$ and $(2017,2018)$ where the function is $2016$. As there are infinitely many primes greater than $2016$, there will be infinitely many points where $f(x)=2016$, at least one each side of each of these primes.