Let $f(x) = \displaystyle\lim_{n\to\infty}\left[\frac{12x\,n^2 + 8x^2 - 1}{n^2 + 10x^2}\right]$, where $[\cdot]$ denotes the greatest integer function. Then the number of positive integers $m$ which satisfy $f\!\left(\dfrac{1}{m}\right) - \dfrac{12}{m} = 0$ is:
Step-by-Step Solution
Key Concept: As $n\to\infty$: $\frac{12xn^2 + 8x^2-1}{n^2+10x^2} \to 12x$. So $f(x) = [12x]$ (the GIF of $12x$). The equation becomes $[12/m] = 12/m$, i.e., $12/m$ must be a positive integer, so $m$ divides $12$.
Divisors of 12: $m \in \{1,2,3,4,6,12\}$ give $12/m \in \{12,6,4,3,2,1\}$ — all integers. But check GIF boundary: for $12/m$ integer, $f(1/m) = 12/m$ only if the limit approaches from above, which holds when $1/m > 1/2$ or $1/m < 1/5$, i.e., $m > 5$ or $m < 2$, giving $m \in \{1, 6, 12\}$. Thus $\mathbf{3}$ values.
Correct Answer: 3