Question

If \(n\gt 0\) and \[ \lim_{x\to 0}\frac{((a-n)nx-\tan x)\sin nx}{x^2}=0, \] then minimum value of \(a\) is

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(-1\)

Hint:

For limits as \(x\to 0\), use \(\tan x\sim x\) and \(\sin nx\sim nx\). After simplifying, apply AM-GM inequality when an expression has the form \(n+\frac{1}{n}\), where \(n\gt 0\).

Answer 1

The Correct Option is B

Step 1: Use standard limits near \(x=0\).
As \[ x\to 0, \] we know that \[ \tan x \sim x \] and \[ \sin nx \sim nx \]

Step 2: Substitute the approximations in the limit.
The given limit is \[ \lim_{x\to 0}\frac{((a-n)nx-\tan x)\sin nx}{x^2}=0 \] Using \[ \tan x\sim x \] and \[ \sin nx\sim nx, \] we get \[ \lim_{x\to 0}\frac{\left((a-n)nx-x\right)(nx)}{x^2}=0 \] \[ = \lim_{x\to 0}\frac{x\left((a-n)n-1\right)nx}{x^2} \] \[ = n\left((a-n)n-1\right) \] Since the limit is \(0\), \[ n\left((a-n)n-1\right)=0 \]

Step 3: Use \(n\gt 0\).
Since \[ n\gt 0, \] we must have \[ (a-n)n-1=0 \] So, \[ (a-n)n=1 \] \[ a-n=\frac{1}{n} \] \[ a=n+\frac{1}{n} \]

Step 4: Find the minimum value of \(a\).
For \[ n\gt 0, \] we know by AM-GM inequality, \[ n+\frac{1}{n}\geq 2 \] Equality holds when \[ n=1 \] Therefore, \[ a_{\min}=2 \]

Step 5: Final conclusion.
Hence, the minimum value of \(a\) is \[ \boxed{2} \]