Step 1: State the given limit and goal.
We are given $n > 0$ and \[\lim_{x\to 0}\frac{\bigl((a-n)nx - \tan x\bigr)\sin nx}{x^2} = 0\] and we must find the minimum value of $a$.
Step 2: Apply small-angle approximations.
Near $x = 0$, two key standard limits are: $\dfrac{\tan x}{x} \to 1$ (so $\tan x \approx x$) and $\dfrac{\sin nx}{nx} \to 1$ (so $\sin nx \approx nx$). Substituting these approximations into the expression: \[\frac{\bigl((a-n)nx - x\bigr)(nx)}{x^2}\]
Step 3: Simplify the resulting expression.
Factor $x$ out of the bracket: $(a-n)nx - x = x[(a-n)n - 1]$. So: \[\frac{x[(a-n)n-1] \cdot nx}{x^2} = n[(a-n)n - 1]\] This is a constant (not involving $x$), so the limit is $n[(a-n)n-1]$.
Step 4: Set the limit equal to zero.
\[n[(a-n)n - 1] = 0\] Since $n > 0$, we cannot have $n = 0$. Therefore: \[(a-n)n = 1 \implies a-n = \frac{1}{n} \implies a = n + \frac{1}{n}\]
Step 5: Minimise $a = n + \frac{1}{n}$ for $n > 0$.
By the AM-GM inequality: for any positive real number $n$, \[n + \frac{1}{n} \geq 2\sqrt{n \cdot \frac{1}{n}} = 2\] Equality holds when $n = 1$.
Step 6: Verify the minimum.
At $n = 1$: $a = 1 + 1 = 2$. Check: $(a-n)n = (2-1)(1) = 1$. The limit $= n \cdot [(a-n)n - 1] = 1 \cdot 0 = 0$.
Step 7: State the final answer.
\[ \boxed{2} \]