Let $f$ be a continuous function from $\mathbb R$ to $\mathbb R$ (where $\mathbb R$ 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) $\lim_\limits{x→∞} f(x)$.
(b) The number of solutions to the equation $f(x) = 2016$.
Is the following reasoning correct for the a part? ' As there are infinite primes, if we assume that ∞ is a prime number, then $f(x)$= $∞$, thus $\lim_{x→∞}$ does not exist.'
In the b part the answer can be obtained by using intermediate value theorem, but shouldn't $f(x)$ not be equal to $2016$ as it is a composite number?
Here's the link to the original question Question Paper
Part A - 9th question.
(a) Your argument is not ok. $\infty$ is not a prime. However, there are arbitrarily large $x$ where $f(x)=2$ and arbitrarily large $x$ where $f(x)=3$ (think of $x=2^n$ and $x=3^n$, for example). Hence $\lim_{x\to\infty}f(x)$ cannot exist.
(b) Between $x_n=2017^n$ where $f(x_n)=2017$ and the even number $x_n+1$ where $f(x_n+1)=2$, the continuity of $f$ guarantees the existence of at least one point $y_n$ where $f(y_n)=2016$. Hence there are infinitely many solutions.